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DIFFERENTIAL EQUATIONS
An equation involving one dependent variable, one or more independent variables and the
differential coefficients (derivatives) of dependent variable with respect to independent variables is
called a differential equation.
Order of a Differential Equation: The order of the highest derivative involved in an ordinary
differential equation is called the order of the differential equation.
Degree of a Differential Equation: The degree of the highest derivative involved in an ordinary
differential equation, when the equation has been expressed in the form of a polynomial in the
highest derivative by eliminating radicals and fraction powers of the derivatives is called the degree
of the differential equation.
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Very Short Answer Questions
1.
Find the order of the family of the differential equation obtained by eliminating the
arbitrary constants b and c from xy = ce x + be − x + x 2 .
Sol.
Equation of the curve is xy = ce x + be − x + x 2
Number of arbitrary constants in the given curve is 2.
Therefore, the order of the corresponding differential equation is 2.
2. Find the order of the differential equation of the family of all circles with their centres at
the origin.
Given family of curves is x 2 + y 2 = a 2 − − − (1) , a parameter.
Diff (1) w.r.t x,
2x+2y.y1 =0.
Hence required differential equation is x+y.y1 = 0.
Order of the differential equation is 1.
 d 2 y  dy 3 
3. Find the order and degree of  2 +   
 dx  dx  


 d 2 y  dy 3 
Sol. Given equation is:  2 +   
 dx  dx  


6/5
= 6y .
6/5
= 6y
3
d 2 y  dy 
i.e. 2 +   = (6y)5 / 6
dx  dx 
Order = 2, degree = 1
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Short Answer Questions
1. Form the differential equation of the following family of curves where parameters are
given in brackets.
i).
y = c ( x − c) ;(c)
2
y = c ( x − c ) − − − − − − (1)
2
Diff. w.r.t x,
y1 = c.2 ( x − c ) − − − − ( 2 )
(1)
⇒
( 2)
y x−c
=
y1
2
⇒ x−c =
2y
y1
and c =x-
2y
y1
2
 2 y  2 y 
from (1) , y =  x-   ⇒ y13 = 4 y ( xy1 − 2 y )
 y1  y1 
ii) xy = ae x + be− x ; ( a, b )
xy = ae x + be − x − − − (1)
Diff. w.r.t. x,
y + x. y1 = ae x − be − x − − − ( 2 )
diff .w.r.t. x,
y1 + y1 + xy2 = ae x + be − x = xy
∴ 2y1 + xy2 = xy
Which is required differential equation.
iii) y = ( a + bx ) e kx ; ( a, b )
y = ( a + bx ) ekx − − − (1)
Diff. w.r.t x,
⇒ y1 = k ( a + bx ) e kx + be kx
⇒ y1 = ky + be kx − − − ( 2 )
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Diff. w.r.t. x,
⇒ y2 = ky1 + kbekx
⇒ y2 = ky1 + k ( y1 − ky )
⇒ y2 = 2ky1 − k 2 y Which is required differential equation.
iv) y = a cos(nx + b); ( a, b )
Ans. y2 + n 2 y = 0
2. Obtain the differential equation which corresponds to each of the following family of
curves.
i) The rectangular hyperbolas which have the coordinates axes as asymptotes.
Sol. Equation of the rectangular hyperbola is xy=c2 where c is arbitrary constant.
Differentiating w.r.t. x
x
dy
+y=0
dx
ii) The ellipses with centres at the origin and having coordinate axes as axes.
Sol. Equation of ellipse is
x2
a2
+
y2
b2
=1
Diff. w.r.t. x,
2x
a2
+
2y dy
b2
=
0
⇒
y.y
=
−
x
1
b 2 dx
a2
Diff. w.r.t. x,
y.y 2 + y1.y1 = −
b2
a2
⇒ y.y2 + 2y1 =
y.y1
x
⇒ x ( y.y2 + 2y1 ) = y.y1
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3. Form the differential equation corresponding to the family of circles of radius r given by
(x – a2) + (y – b)2 = r2, where a and b are parameters.
Sol. We have: (x – a2) + (y – b)2 = r2
…(1)
Differentiating (1) w.r.to x
2(x − a) + 2(y − b)
dy
=0
dx
… (2)
Differentiating (2) w.r.to x
2
1 + (y − b)
d 2 y  dy 
+  = 0
dx 2  dx 
From (2) (x − a) = −(y − b)
… (3)
dy
dx
Substituting in (1), we get
2  dy 
2
(y − b)   + (y − b)2 = r 2
 dx 
  dy 2  2
(y − b)    + 1 = r
  dx 



2
… (4)
  dy  2 
From (3) (y − h) 2 = −  1 +   
  dx  
dx


d2 y
  dy 2 
1 +   
  dx  

(y − h) = − 
2
d y
 2 
 dx 
3
  dy 2 
1 +   
  dx  
 = r2
Substituting in (4): 
2
 d2y 
 2 
 dx 
2
 d 2 y    dy 2 
i.e. r 2  2  = 1 +   


 dx    dx  
3
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Long Answer Questions
1. Form the differential equations of the following family of curves where parameters are
given in brackets.
i)
y = ae3x + be4x ; (a, b)
Sol. y = ae3x + be4x − − − − (1)
Differentiating w.r.to x
y1 = 3ae3x + 4be4x − − − − − ( 2 )
Differentiating w.r.to x,
y 2 = 9ae3x + 16be4x − − − − − ( 3)
Eliminating a,b from above equations,
y
e3 x
e4 x
y1
3e3 x
4e 4 x = 0
y2
9e3 x
16e4 x
y
1
1
⇒ y1
3
4 =0
y2
9
16
⇒ y2 − 7 y1 + 12 y = 0 Which is the required differential equation.
ii) y = ax2 + bx ; (a, b)
Sol.
y = ax2 + bx
---- (1)
diff. w.r.t. x,
y1 = y 2 x + b ⇒ y1x = y 2 x 2 + bx − − − ( 2 )
diff. w.r.t. x,
y 2 = 2a
----------- (3)
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From (2) and (3),
y1 = y 2 x + b ⇒ y1x = y 2 x 2 + bx − − − ( 4 )
x
2
d2 y
dx
−2x
2
= 2ax 2
… (i)
dy
= −4ax 2 − 2bx
dx
2y = 2ax2 + 2bx
… (ii)
… (iii)
Adding all three equations we get
x2
d2 y
dx
2
− 2x
dy
+ 2y = 0
dx
iii) ax2 + by2 = 1 ; (a, b)
Sol.
Given equation is
ax2 + by2 = 1
------ (1)
Differentiating w.r.t. x
⇒ 2ax + 2byy1 = 0
⇒ ax + byy1 = 0 − − − − − − ( 2 )
Differentiating w.r.t. x
(
)
⇒ a + b ( yy2 + y1 y1 ) = 0 ⇒ a + b yy2 + y12 = 0
(
)
⇒ ax + bx yy2 + y12 = 0 ---- (3)
( 3) − ( 2 ) ⇒ bx ( yy2 + y12 ) − byy1 = 0
(
)
⇒ x yy2 + y12 − yy1 = 0
b
x
iv) xy = ax 2 + ; (a, b)
Sol. xy = ax 2 +
b
x
x2y = ax2 + b
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Differentiating w.r.t. x
x2y1 + 2xy = 3ax2
Dividing with x
xy1 + 2y = 3ax
… (i)
Differentiating w.r.t. x
xy2 + y1 + 2y1 = 3a
xy2 + 3y1 = 3a
… (ii)
Dividing (i) by (ii)
xy1 + 2y 3ax
=
=x
xy 2 + 3y1 3a
Cross multiplying
xy1 + 2y = x 2 y 2 + 3xy
x 2 y 2 + 2xy1 − 2y = 0
 d2 y 
 dy 
x 2  2  + 2x   − 2y = 0
 dx 
 dx 


2. Obtain the differential equation which corresponds to each of the following family of
curves.
i)
The circles which touch the Y-axis at the origin.
Sol. Equation of the given family of circles is
x2 + y2 + 2gx = 0, g arbitrary const …(i)
x2 + y2 = –2gx
Differentiating w.r.t. x
2x + 2yy1 = –2g
… (ii)
Substituting in (i)
x2 + y2 = x(2x + 2yy1) by (ii)
= 2x2 + 2xyy1
yy2 – 2xyy1 – 2x2 = 0
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y2 – x2 = 2xy
dy
.
dx
ii) The parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis.
Sol.
Equation of the given family of parabolas is
(y – k)2 = 4a(x – h)
-----(i)
Where h,k are arbitrary constants
Differentiating w.r.t. x
2(y – k)y1 = 4a
(y – k)y1 = 2a
…(2)
Differentiating w.r.t. x
(y – k)y2 + y12 = 0
From (2), y – k =
…(3)
2a
y1
Substituting in (3)
2a
⋅ y 2 + y12 = 0 ⇒ 2ay 2 + y13 = 0
y1
iii) The parabolas having their foci at the origin and axis along the x-axis.
Sol.
Given family of parabolas is y2 = 4a(x + a) ----- (i)
Diff. w.r.t.x,
2y
dy
1
= 4a ⇒ yy′ = a
dx
2
----- (2)
From (i) and (2),
y2 = 4
1
1


yy′  x +
yy′ 
2
2


1
y 2 = 2y′x + 4 ⋅ y 2 y′2 ⇒ y 2 = 2yy′x + y 2 y′2
4
2
 dy 
 dy 
y   + 2x   = y
 dx 
 dx 
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Solutions of Differential Equations
Variables Separable:
Let the given equation be
dy
= f (x, y) . If f(x, y) is a variables separable function,
dx
i.e., f(x, y) = g(x)h(y) then the equation can be written as
integrating both sides, we get the solution of
dy
dy
= g (x ) h ( y) ⇒
= g(x)dx . By
dx
h(y)
dy
= f (x, y) . This method of finding the solution is
dx
known as variables separable.
Very Short Answer Questions
1. Find the general solution of 1 − x 2 dy + 1 − y 2 dx = 0 .
Sol. Given d.e. is
1 − x 2 dy + 1 − y 2 dx = 0
1 − x 2 dy = − 1 − y 2 dx
Integrating both sides
∫
dy
1 − y2
= −∫
dx
1− x2
sin–1y = –sin–1x + c
Solution is sin–1x + sin–1y = c, where c is a constant.
2. Find the general solution of
Sol.
dy 2y
=
.
dx x
dy 2y
dy
dx
=
⇒∫
=2
dx x
y
x
Integrating both sides
log c + log y = 2 log x
log cy = log x 2
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Solution is cy = x2 where c is a constant.
Short Answer Questions
Solve the following differential equations.
1.
dy 1 + y2
=
dx 1 + x 2
Sol.
dy 1 + y2
=
dx 1 + x 2
Integrating both sides
⇒∫
dy
1 + y2
=∫
dx
1+ x2
tan −1 y = tan −1 x + tan −1 c Where c is a constant.
2.
dy
= e y− x
dx
dy e y
dy dx
Sol.
= x ⇒ y = x
dx e
e
e
Integrating both sides ∫ e − x dx = ∫ e − y dy ⇒ −e− x = −e− y + c
e− y = e− x + c Where c is a constant.
3. (ex + 1)y dy + (y + 1)dx = 0
Sol. (ex + 1)y dy = –(y + 1)dx
ydy
dx
=− x
y +1
e +1
Integrating both sides

1 
e− x dx
1
−
dy
=
−
∫  y + 1  ∫ e− x + 1
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y − log(y + 1) = log(e− x + 1) + log c
⇒ y − log(y + 1) = log c(e − x + 1)
⇒ y = log(y + 1) + log c(e− x + 1)
y = log c(y + 1)(e − x + 1)
Solution is: e y = c(y + 1)(e − x + 1) .
4.
dy
= e x − y + x 2e− y
dx
dy
ex x 2
x −y
2 −y
=e
+x e = y + y
Sol.
dx
e
e
Integrating both sides
∫e
y
⋅ dy = ∫ (e x + x 2 )dx
Solution is: e y = e x +
x3
+c
3
5. tan y dx + tan x dy = 0
Sol. tan y dx = –tan x dy
dx
−dy
cos x
cos y
=
⇒
dx = −
dy
tan x tan y
sin x
sin y
Taking integration
cos x
cos y
∫ sin x dx = −∫ sin y dy
log sin x = − log sin y + log c
log sin x + log sin y = log c
log(sin x ⋅ sin y) = log c ⇒ sin x ⋅ sin y = c
6.
1 + x 2 dx + 1 + y 2 dy = 0
Sol. 1 + x 2 dx = − 1 + y 2 dy
Integrating both sides
∫
1 + x 2 dx = − ∫ 1 + y 2 dy
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x
1
× 1 + x 2 + sinh −1 x =
2
2
y
1 + y2 1
= sinh −1 x + c
2
2
x 1 + x 2 + y 1 + y 2 + log (x + 1 + x 2 )(y + 1 + y 2 )  = c


7.
y−x
dy
dy 

= 5  y2 + 
dx
dx 

Sol. y − 5y 2 = (x + 5)
dy
dx
dy
⇒
=
dx
x + 5 y(1 − 5y)
Integrating both sides
dx
1
dy

5
∫ x + 5 = ∫ y(1 − 5y) = ∫  y + 1 − 5y  dy
ln | x + 5 |= ln y − ln |1 − 5y | + ln c
ln | x + 5 |= ln
 cy 
cy
⇒ x +5 = 

1 − 5y
 1 − 5y 
8.
dy xy + y
=
dx xy + x
Sol.
dy y(x + 1)
y +1
x +1
=
⇒
dy =
dx
dx x(y + 1)
y
x

1

1
∫ 1 + y  dy = ∫ 1 + x  dx
y + log y = x + log x + log c
y − x = log
cx
y
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Short Answer Questions
1.
dy
1 + y2
=
dx (1 + x 2 )xy
dy
1 + y2
=
dx (1 + x 2 )xy
Sol.
ydy
dx
⇒
=
2
1+ y
x(1 + x 2 )
2ydy
1 + y2
=
2xdx
x 2 (1 + x 2 )
Integrating both sides
2ydy
 1

1
∫ 1 + y2 = ∫  x 2 − 1 + x 2  2x dx
log(1 + y 2 ) = log x 2 − log(1 + x 2 ) + log c
log(1 + x 2 ) + log(1 + y 2 ) = log x 2 + log c
Solution is: (1 + x2)(1 + y2) = cx2
2.
dy
+ x 2 = x 2 ⋅ e3y
dx
Sol.
dy
+ x 2 = x 2 ⋅ e3y
dx
⇒
dy
= x 2 ⋅ e3y − x 2 = x 2 (e3y − 1)
dx
Integrating both sides
dy
∫ e3y − 1 = ∫ x
log
2
dx ⇒ ∫
e−3y
1− e
−3y
= ∫ x 2dx
(1 − e−3y ) x 3
=
+c
3
3
log(1 − e −3y ) = x 3 + c′ (c′ = 3c)
Solution is: 1 − e −3y = e x ⋅ k (k = ec′ )
3
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3. (xy2 + x)dx + (yx2 + y)dy = 0
Sol. (xy2 + x)dx + (yx2 + y)dy = 0
x(y2 + 1)dx + y(x2 + 1)dy = 0
Dividing with (1 + x2)(1 + y2)
xdx
1+ x2
+
ydy
1 + y2
=0
Integrating both sides
xdx
ydy
∫ 1 + x 2 + ∫ 1 + y2 = 0
1
(log(1 + x 2 ) + log(1 + y 2 )  = log c


2
log(1 + x 2 )(1 + y 2 ) = 2 log c = log c2
(1 + x2)(1 + y2) = k when k = c2.
4.
dy
= 2y tanh x
dx
Sol.
dy
dy
= 2y tanh x ⇒
= 2 tanh xdx
dx
y
Integrating both sides
∫
dy
= 2 ∫ tanh x dx
y
log y = 2 log cosh x + log c
ln y = 2 ln cosh x + ln c ⇒ y = c cos 2 hx
5.
 dy 
sin −1   = x + y
 dx 
dy
= sin(x + y) ⇒ x + y = t
dx
1+
dy dt
=
dx dx
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dt
dt
− 1 = sin t ⇒
= 1 + sin t
dx
dx
dt
= dx
1 + sin t
Integrating both sides
dt
∫ 1 + sin t = ∫ dx
∫
1 − sin t
cos 2 t
∫ sec
2
dt = x + c
tdt − ∫ tan t ⋅ sec t dt = x + c
tan t − sec t = x + c
⇒ tan(x + y) − sec(x + y) = x + c
6.
dy y 2 + y + 1
+
=0
dx x 2 + x + 1
−dy
y + y +1
2
=
dx
x + x +1
2
Integrating both sides
−∫
dy
y2 + y + 1
dy
=∫
dx
x2 + x + 1
dx
−∫
=
∫
2
2
1 3
1 3


y
+
+
x
+



 +
2 4
2 4


2
(y + 1/ 2)
2
(x + 1/ 2)
tan −1
=
tan −1
+c
3
3/ 2
3
3/ 2
2x + 1
2y + 1
tan −1
+ tan −1
=c
3
3
−
7.
dy
= tan 2 (x + y)
dx
Sol.
dy
= tan 2 (x + y) put v = x + y
dx
dv
dy
= 1+
= 1 + tan 2 v = sec2 v
dx
dx
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dv
∫ sec2 v = ∫ dx = ∫ cos
2
v ⋅ dv = x + c
(1 + cos 2v)
dv = x + c
2
⇒ ∫ (1 + cos 2v)dv = 2x + 2c
∫
sin 2v
= 2x + 2c
2
2v + sin 2v = 4x + c′
v+
2(x + y) + sin 2(x + y) = 4x + c′
1
x − y − sin[2(x + y)] = c
2
Homogeneous Equations
Homogeneous Differential Equations:
A differential equation
dy f (x, y)
=
is said to be a homogeneous differential equation in x, y if both
dx g(x, y)
f(x, y), g(x, y) are homogeneous functions of same degree in x and y.
To find the solution of the h .d.e put y = vx, then
dy
dv
= v+x
. Substituting these values in given
dx
dx
differential equation, then it reduces to variable separable form. Then we find the solution of the
d.e.
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Very Short Answer Questions
 y
dy
1. Express xdy − ydx = x 2 + y2 dx in the form F   =
.
 x  dx
Sol. x ⋅ dy − ydx = x 2 + y 2 dx
x
dy
dy
− y = x 2 + y2 ⇒ x
= y + x 2 + y2
dx
dx
dy y
x 2 + y2 y
 y
= +
= + 1+  
2
dx x
x
x
x
 y
2
dy
Which is of the form F   =
 x  dx

y
y
 y
dy
2. Express  x − yTan −1  dx + x tan −1 dy = 0 in the form F   =
.
x
x
 x  dx



y
y
Sol. Given  x − yTan −1  dx + x tan −1 dy = 0
x
x

y
y

x tan −1   dy = −  x − y tan −1  dx
x
x

y
 y  dy
 y
tan −1  
= −  1 − tan −1 
x
 x  dx
 x
=
y
y
tan −1   − 1
x
x
y
 y
⋅ tan −1   − 1
dy x
 x  = F y 
=
 
dx
y
x
tan −1  
x
3. Express x ⋅
Sol. x ⋅
dy
 y  dy
= y(log y − log x + 1) in the form F   =
.
dx
 x  dx
dy
= y(log y − log x + 1)
dx
dy y 
y 
=  log + 1
dx x 
x 
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Short Answer Questions
Solve the following differential equations.
1.
dy x − y
=
dx x + y
Sol.
dy x − y
=
----- (1)
dx x + y
(1) is a homogeneous d.e.
Put y = vx
dy
dv
= v+x
dx
dx
dv x − vx x(1 − v)
v+x
=
=
dx x + vx x(1 + v)
dv 1 − v
1 − v − v − v 2 1 − 2v − v 2
=
−v=
=
x⋅
dx 1 + v
1+ v
1+ v
(1 + v)dv
∫ 1 − 2v − v2 = ∫
dx
x
1
− log(1 − 2v − v 2 ) = log x + log c
2

1
y y2 
− log  1 − 2 ⋅ − 2  = log cx

2
x x 

log
(x 2 − 2xy − y2 )
x
2
x 2 − 2xy − y 2
x
2
= −2log cx = log(cx) −2
= (cx) −2 =
(x 2 − 2xy − y 2 ) =
1
c2
1
2 2
c x
=k
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2. (x2 + y2)dy = 2xy dx
Sol.
dy
2xy
= 2
which is a homogeneous d.e.
dx x + y 2
Put y = vx
dy
dv
= v+x⋅
dx
dx
v+x⋅
x⋅
dv
2x(vx)
2v
= 2
=
dx x + v 2 x 2 1 + v 2
dv
2v
2v − v − v3 v − v3
=
−
v
=
=
dx 1 + v 2
1 + v2
1 + v2
1 + v2
∫ v(1 − v2 ) dv = ∫
Let
1 + v2
v(1 − v )
2
=
dx
x
A
B
C
+
+
v 1+ v 1− v
1 + v 2 = A(1 − v 2 ) + BV(1 − v) + CV(1 + v)
1 + v2
dv
dv
dv
v = 0 ⇒1= A
dv = ∫ − ∫
+∫
∫
2
v
1+ v
1− v
v(1 − v )
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
v = −1 ⇒ 1 + 1 = B(−1)(2) ⇒ 2 = −2B ⇒ B = −1
= log v − log(1 + v) − log(1 − v) = log
∴ log
v
1 − v2
v
1− v
2
v
1 − v2
= log x + log c = log cx
= cx ⇒ v = cx(1 − v 2 )
 y2 
y
y
(x 2 − y 2 )
= cx 1 − 2  ⇒ = cx
 x 
x
x
x2


Solution is: y = c(x 2 − y 2 )
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3.
dy −(x 2 + 3y 2 )
=
dx (3x 2 + y2 )
Sol.
dy −(x 2 + 3y 2 )
=
which is a homogeneous d.e.
dx (3x 2 + y2 )
Put y = vx
dy
dv
= v+ x⋅
dx
dx
v+ x⋅
x⋅
dv −(x 2 + 3v 2 x 2 ) − x 2 (1 + 3v 2 )
=
= 2
dx
3x 2 + v 2 x 2
x (3 + v 2 )
dv
1 + 3v 2
= −v −
dx
3 + v2
=
−3v − v3 − 1 − 3v 2
3 + v2
3 + v2
(v + 1)
3
3 + v2
(v + 1)
3
=−
(v + 1)3
3 + v2
=
−dx
x
=
A
B
C
+
+
2
v + 1 (v + 1)
(v + 1)3
Multiplying with (v + 1)3
3 + v2 = A(v + 1)2 + B(v + 1) + C
v = –1 ⇒ 3 + 1 = C ⇒ C = 4
Equating the coefficients of v2
A=1
Equating the coefficients of V
0 = 2A + B
B = –2A = –2
v2 + 3
(v + 1)
3
=
v2 + 3
1
2
4
−
+
2
v + 1 (v + 1)
(v + 1)3
∫ (v + 1)3 = −∫
dx
x
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 1
2
4

2
4
c
∫  v + 1 − (v + 1)2 + (v + 1)3  dv = − log x + log c log(v + 1) + v + 1 − 2(v + 1)2 = log x


Solution is:
2
2
c
y 
log  + 1 +
−
= log
2
x
 x  y +1  y 
1
+


x
x 
2x
2x 2
c
(x + y)
−
= log − log
2
x + y (x + y)
x
x
2x 2 + 2xy − 2x 2
(x + y)
2xy
(x + y)
2
2
= log
= log
c
x+y
c
x+y
 c 
2xy
x+y
log 
=−
 c = − log 
(x + y) 2
 c 
x+y
4. y2 dx + (x2 – xy)dy = 0
Sol. y2 dx + (x2 – xy)dy = (xy – x2)dy
dy
y2
=
Which is a homogeneous d.e.
dx xy − x 2
Let y = vx ⇒
v+x
x⋅
dy
dv
= v+x⋅
dx
dx
dv
v2 x 2
= 2
dx x (v − v2 )
dv
v2
v2 − v2 + v
=
−v=
dx v − 1
v −1
v −1
dx
dx
 1
dv =
⇒ ∫ 1 −  dv = ∫
v
x
x
 v
v − log v = log x + log k
v = log v + log x + log k = log k(vx)
y
= log ky ⇒ ky = e y / x
x
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dy (x + y) 2
=
dx
2x 2
5.
dy (x + y) 2
=
Which is a homogeneous d.e.
dx
2x 2
Sol,.
y = vx ⇒
v+x
x
dy
dv
= v+x
dx
dx
2
dv (x + vx)2
2 (1 + v)
=
=
x
dx
2x 2
2x 2
dv (1 + v 2 )
1 + v 2 + 2v − 2v
=
−v=
dx
2
2
2∫
dv
1 + v2
=∫
dx
⇒ 2 tan −1 v = log x + log c
x
 y
2 tan −1   = log cx
x
6. (x2 – y2)dx – xy dy = 0
Sol. (x2 – y2)dx – xy dy = 0
(x2 – y2)dx = xy dy
dy x 2 − y 2
=
dx
xy
dy
dv
= v+x
y = vx ⇒
dx
dx
v+x
x
dv x 2 − v 2 x 2 x 2 (1 − v 2 )
=
=
dx
vx 2
vx 2
dv 1 − v 2
1 − v 2 − v 2 1 − 2v 2
=
−v=
=
dx
v
v
v
vdv
∫ 1 − 2v2 = ∫
dx
x
1
− log(1 − 2v 2 ) = log x + log c
4
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 2y 2 
1
− log  1 − 2  = log x + log c

4
x 

 x 2 − 2y 2 
1
− log 
 = log x + log c
 x2
4


1
− log(x 2 − 2y 2 ) − log x 2  = log x + log c

4
1
1
− log(x 2 − 2y 2 ) + ⋅ 2 log x = log x + log c
4
4
1
1
− log(x 2 − 2y 2 ) = log x + log c
4
2
− log(x 2 − 2y 2 ) = −2 log x − 4 log c
log(x 2 − 2y2 ) = −2 log x + log k where k =
1
c4
= log
k
x2
⇒ x 2 − 2y 2 =
Solution is: x2(x2 – 2y2) = k
7. (x2y – 2xy2)dx = (x3 – 3x2y)dy
Sol. (x2y – 2xy2)dx = (x3 – 3x2y)dy
dy x 2 y − 2xy 2
Which is a homogeneous d.e.
= 3
dx
x − 3x 2 y
Put y = vx so that
v+x
dv x 3 v − 2v 2 x 3
= 3
dx
x − 3vx 3
=
x
dy
dv
= v+x
dx
dx
(v − 2v 2 )x 3
(1 − 3v)x 3
v − 2v 2
=
1 − 3v
dv v − 2v 2
=
−v
dx 1 − 3v
=
v − 2v 2 − v(1 − 3v) −2v 2 + 3v 2
=
1 − 3v
1 − 3v
dv
v2
1 − 3v
dx
x
=
⇒ 2 dv =
dx 1 − 3v
x
v
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k
x2
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 1
3
∫  v2 − v  dv = ∫
dx
x
−1
− 3log v = log x + log c
v
−x
y
= 3log   = log x + log c
y
x
3
−x
y
− log   = log xc
y
x
−x
y3
= log xc + log 3
y
x

 cy3 
−x
y3 
= log  cx ⋅ 3  = log  2 

 x 
y
x 



cy3
x2
= e − x / y ⇒ cy3 =
x2
ex / y
cy3 ⋅ e x / y = x 2
8. y2 dx + (x2 – xy + y2)dy = 0
Sol. y2 dx = –(x2 – xy + y2)dy
dy
− y2
= 2
dx x − xy + y 2
dy
dv
y = vx ⇒
= v+x
dx
dx
v+x
x
=
dv
−v2 x 2
− v2 x 2
= 2
=
dx x − vx 2 + v 2 x 2 x 2 (1 − v + v 2 )
dv
− v2
=
−v
dx 1 − v + v 2
− v 2 − v + v 2 − v3
1 − v + v2
1 − v + v2
v(1 + v 2 )
Let
dv = −
1 − v + v2
v(1 + v )
2
=
=−
v(1 + v 2 )
1 − v + v2
dx
x
… (1)
A Bv + C
+
v 1 + v2
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1 − v + v2 = A(1 + v 2 ) + (Bv + C)v
v=0⇒1=A
Equating the coefficients of v2
1=A+B⇒B=0
Equating the coefficients of v
−1 = C ⇒
1 − v + v2
v(1 + v )
2
1 − v + v2
∫ v(1 + v2 ) dv = ∫
=
1
1
−
v 1 + v2
dv
dv
−∫
= log v − tan −1 v From (1) we get
2
v
1+ v
log v − tan −1 v = − log x + log c
tan −1 v = log v + log x − log c = log
vx
y
= log
c
c
−1
−1
y
= e tan v = e tan (y / x)
c
Solution is: y = c ⋅ e tan
−1
(y / x)
9. (y2 – 2xy)dx + (2xy – x2)dy = 0
Sol. (y2 – 2xy)dx + (2xy – x2)dy = 0
(2xy – x2)dy = –(y2 – 2xy)dx
dy 2xy − y2
=
dx 2xy − x 2
Put y = vx ⇒
dy
dv
= v+x
dx
dx
dv 2vx 2 − v2 x 2 x 2 (2v − v 2 )
=
= 2
v+x
dx
2vx 2 − x 2
x (2v − 1)
x
dv 2v − v 2
=
−v
dx
2v − 1
=
2v − v 2 − 2v 2 + v 3v(1 − v)
=
2v − 1
2v − 1
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2v − 1
∫ v(1 − v) dv = 3∫
dx
x
… (1)
2v − 1 A
B
= +
v(1 − v) v 1 − v
2v − 1 = A(1 − v) + Bv
Let
v = 0 ⇒ −1 = A ⇒ A = −1
v =1⇒1= B ⇒ B =1
 1
1 
dx
x
− log v − log(1 − v) = 3log x + log c
∫  − v + 1 + v  dv = 3∫
log
1
= log cx 3
v(1 − v)
1
1
= cx 3 ⇒ v(1 − v) = 3
v(1 − v)
cx
y y
1
yx−y
1
1−  = 3 ⇒ 
= 3


x  x  cx
x  x  cx
1
1
xy(x − y) = = k ⇒ xy(y − x) = − = k
c
c
10.
dy y y 2
+ =
dx x x 2
Sol.
dy y y 2
+ =
Which is a homogeneous d.e.
dx x x 2
Put y = vx ⇒
dy
dv
= v+x
dx
dx
dv
v2 x 2
dv
v+x
+v= 2 ⇒x
= v2 − 2v
dx
dx
x
dv
v − 2v
2
Let
=
dy
x
1
v − 2v
2
=
A
B
+
v v−2
1 = A(v − 2) + Bv
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1
2
1
v = 2 ⇒ 1 = 2B ⇒ B =
2
v = 0 ⇒ 1 = A(−2) ⇒ −
−
1 1
1 
dx
 −
 dv = ∫
∫
2  v v−2
x
−
1
[log v − log(v − 2)] = log x + log c
2
1
v 
− log
= log cx
2
v − 2 
log
v
= − log cx = log(cx)−2
v−2
v
(y / x)
1
= (cx)−2 ⇒
= 2 2
v−2
(y / x) − 2 c x
y
1
1
= 2 2 ⇒ x 2 y = 2 (y − 2x)
y − 2x c x
c
Solution is:
y − 2x = c2 x 2 y = kx 2 y Where k = c2
11. xdy − ydx = x 2 + y2 dx
Sol. xdy − ydx = x 2 + y2 dx
x
dy
− y = x 2 + y2
dx
dy y
− =
dx x
x 2 + y2
x
Put y = vx ⇒
dv
∴x =
dx
∫
dv
1 + v2
dy
dv
= v+x
dx
dx
x 2 + y 2 x 1 + v2
=
x
x
=∫
dx
⇒ sinh −1 v = log x + log c
x
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log  v + 1 + v 2  = log cx ⇒ v + 1 + v 2 = cx


y
y2
+ 1 + 2 = cx ⇒ y + x 2 + y 2 = cx 2
x
x
12. (2x – y)dy = (2y – x)dx
Sol.
dy 2y − x
=
dx 2x − y
Put y = vx ⇒
dv x(2v − 1)
=
dx x(2 − v)
v+x
x
dy
dv
= v+x
dx
dx
dv 2v − 1
2v − 1 − 2v + v 2
=
−v=
dx 2 − v
2−v
2−v
v −1
2
dv =
dx
dv
vdv
dx
⇒ 2∫ 2
−∫ 2
=∫
x
x
v −1
v −1
1
v −1 1
2 ⋅ log
− log(v 2 − 1) = log x + log c
2
v +1 2
1
v −1

− log(v 2 − 1)  = log cx
 2 log
2
v +1

1
(v − 1) 2
1
⋅
= log cx
log
2
2
(v + 1) (v − 1)(v + 1)
log
v −1
(v + 1)
2
= 2 log cx = log c 2 x 2
y
−1
2 2
x
c
x
⇒
= c2 x 2
∴
=
3
3
(v + 1)
y 
 + 1
x 
v −1
y−x
x 2 (y − x)
2 2
x
=c x ⇒
= c2 x 2
3
3
(y − x)
(x + y)
x3
(y – x) = c2 (x + y)3.
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13. (x 2 − y 2 )
Sol.
dy
= xy
dx
dy
xy
= 2
dx (x − y 2 )
Put y = vx ⇒
v+x
x
dv
x(vx)
v
= 2
=
2
2
dx x − v x
1 − v2
dv
v
v − v + v3
v3
v
=
−
=
=
dx 1 − v 2
1 − v2
1 − v2
1 − v2
v
−
dy
dv
= v+x
dx
dx
3
1
2v 2
dv =
dx
dv
dv
dx
⇒∫ 3 −∫
=∫
x
v
x
v
− log v = log x + c
1 x2
−
= log vx + c = log y + c
2 y2
−x 2
2y
2
= (log y + c) ⇒ − x 2 = 2y 2 (c + log y)
x2 + 2y2 (c + log y) = 0.
14. Solve 2
dy y y2
= +
dx x x 2
Sol. Put y = vx ⇒
2v + 2x
dy
dv
= v+x
dx
dx
dv
dv
= v + v 2 ⇒ 2x
= v2 − v
dx
dx
dv
dx
1
dx
 1
=2
⇒ ∫
−  dv = 2 ∫
v(v − 1)
x
x
 v −1 v 
log(v – 1) – log v = 2 logx + log c
log
v −1
v −1
= log cx 2 ⇒
= cx 2
v
v
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y
−1
y−x
x
= cx 2 ⇒
= cx 2
y
y
x
Solution is : (y – x) = cx2y
Long Answer Questions
(
)

x


1. Solve 1 + e x / y dx + e x / y 1 −  dy = 0 .
y
(
)

x


Sol. 1 + e x / y dx + e x / y 1 −  dy = 0
y
 x
e x/ y  1 − 
dx
 y  Which is a homogeneous d.e.
⇒
=−
dy
1 + e x/y
(
Put x = vy ⇒
(1 + e v )
)
dx
dv
= v+y
dy
dy
dx v
+ e (1 − v) = 0
dy

dv 
(1 + e v )  v + y  + e v (1 − v) = 0
dy 

dv
v + ve v + y(1 + e v ) + e v − ve v = 0
dy
y(1 + e v )dv = −(v + e v )dy
1 + ev
∫ v + ev dv = − ∫
dy
y
log(v + e v ) = − log y + log c ⇒ v + e v =
c
y
x x/y c
+e
= ⇒ x + y ⋅ ex / y = c
y
y
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y dy
y
= y sin − x
x dx
x
2. Solve x sin ⋅
y dy
y
= y sin − x
x dx
x
Sol. x sin ⋅
y  y x
 sin   − 
dy x   x  y 
⇒
=
dx
y
sin  
x
Put y = vx ⇒
dy
dv
= v+x
dx
dx
1

v  sin v − 
dv
v
v+x
= 
dx
sin v
x
dv v sin v − 1 − v sin v
=
dx
sin v
− sin vdv =
1
dx
x
dx
x
⇒ cos v = log x + log c = log cx
⇒ ∫ − sin v ⋅ dv = + ∫
⇒ cx = ecos v = ecos(y / x) .


y
3. Solve xdy =  y + x cos 2  dx .
x
Sol. x

dy
y
dy y
y
= y + x ⋅ cos 2 ⇒
= + cos 2
dx
x
dx x
x
Put y = vx,
dy
dv
dv
= v+x
⇒ v+x
= v + cos 2 v
dx
dx
dx
dv
dx
dx
2
=
⇒
sec
v
⋅
dv
=
∫
∫
x
cos 2 v x
tan v = logx + c
y
i.e. tan   = log x + c .
x
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4. Solve: (x – y log y + y log x)dx + x(log y – log x)dy = 0.
Sol. Dividing with x. dx we get
1−
y
y
 y  dy
log y + log x + log  
=0
x
x
 x  dx
y
y
 y  dy
1 −  log y − log  + log  
=0
x
x
 x  dx
Put y = vx,
dy
dv
= v+x
dx
dx
dv 

1 − v[log v] + log v  v + x  = 0
dx 

dv
1 − v log v + v log v + x log v
=0
dx
x log v
dv
dx
= −1 ⇒ ∫ log v dv = − ∫
dx
x
v log v − ∫ dv = c − log x
v log v − v = c − log x
v + c = v log v + log x ⇒ v + c = v log v + log x
y
y
 y
+ c = log   + log x
x
x
x
 y
y + cx = y log   + x log x
x
= y log y − y log x + x log x
= (x − y) log x + y log y
5. Solve: (ydx + xdy)x cos
Sol. (ydx + xdy)x cos
y
y
= (xdy − ydx)y sin
x
x
y
y
= (xdy − ydx)y sin
x
x
y
y 
y
y  dy

2
2
=0
 xy ⋅ cos + y sin  −  xy ⋅ sin − x cos 
x
x 
x
x  dx

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y
y
xy ⋅ cos   + y 2 sin  
dy
x
x
=
dx
y
y
xy sin   − x 2 cos  
x
x
2
y
 y  y
y
  cos   +   sin  
x
x x
 x  = F y 
= 
 
y y
 y
x
sin
−
cos
   
 
x x
x
∴ This is a homogeneous equation
y = vx ⇒
v+x
dy
dv
= v+x
dx
dx
dv v cos v + v 2 sin v
=
dx
v sin v − cos v
dv v cos v + v 2 sin v
x
=
−v
dx
v sin v − cos v
v cos v + v 2 sin v − v 2 sin v + v cos v
v sin v − cos v
2v cos v
=
v sin v − cos v
=
v sin v − cos v 2
= dv
v cos v
x

1
2
∫  tan v − v  dv = ∫ x dv
log sec v − log v = 2 log | x |
∴ log x 2 = log
But
c
c
⇒ x2 =
v cos v
v cos v
y
=v
x
Solution is:
x2 =
y
⇒ xy cos   = c
y
y
x
⋅ cos  
x
x
c
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6. Find the equation of a curve whose gradient is
dy y
y
= − cos 2 , where x > 0, y > 0 and
dx x
x
which passes through the point (1, π/4).
Sol.
dy y
y
= − cos 2 which is homogeneous d.eq.
dx x
x
Put y = vx
dy
dv
= v+x
dx
dx
v+x
∫ sec
dv
dv
dx
= v − cos 2 v ⇒ ∫
=
−
∫
dx
x
cos 2 v
2
v = −∫
dx
⇒ tan v = − log | x | + c
x
This curve passes through (1, π/4)
π
tan   = c − log1 ⇒ c = 1
4
Equation of the curve is :
 y
tan v = 1 − log | x |⇒ tan  1 − log | x |
x

x


7. Solve (1 + 2e x / y )dx + 2e x / y 1 −  dy = 0 .
y

x


Sol. (1 + 2e x / y )dx = −2e x / y 1 −  dy
y
x 
= 2e x / y  − 1 dy
y 
x 
2e x / y  − 1
dy
y 
=
dx
1 + 2e x / y
Put x = vy
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dx
dv
dv 2e v (v − 1)
= v+y
⇒ v+y
=
dy
dy
dy
1 + 2e v
y
=
dv 2e v (v − 1)
=
−v
dy
1 + 2e v
2ve v − 2e v − v − 2v
1 + 2e v
1 + 2e v
∫ v + 2e v dv = −∫
e =
v
−(2e v + v)
1 + 2e v
dy
y
log(v + 2e v ) = − log y + log c = log
v + 2e v =
c
y
c
y
Solution is:
x
c
+ 2e x / y = ⇒ x + 2y ⋅ e x / y = c
y
y
 y
 
y
 
Solve x sec   (ydx + xdy) = y csc   (xdy − ydx) .
x
x
8.
 y
 
y
 
 y 
 
dy 
 y  dy
 


Sol. x sec   (ydx + xdy) = y csc   (xdy − ydx) ⇒ x sec   y + x  = y csc   x − y 
x
x
x
dx
x
dx
x
dy 
y
 y 
x ⋅ sec   − y ⋅ csc   

dx 
x
 x 

y
 y 
= − y  y csc   + x sec   
x
 x 


y
 y 
− y  y csc   + x sec   
dy
x
 x 

=
dx

y
 y 
x  x sec   − y csc   
x
 x 

This is a homogeneous equation.
Put y = vx
dy
dv
= v+x
dx
dx
v+x
dv
 v csc v + sec v 
= v

dx
 v csc v − sec v 
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1 
 v
+
v

sin v cos v  v(v cosv + sin v)
= 
=
1
1 
v cos v − sin v

−
v

 sin v cos v 
dv v(v cos v + sin v)
=
−v
dx
v cos v − sin v
v(v cos v + sin v − v cos v + sin v)
=
v cos v − sin v
2v sin v
=
v cos v − sin v
x
v cos v − sin v
dx
dv = 2 ∫
v sin v
x
cos v
1
dx
∫ sin v dv − ∫ v dv = 2∫ x
∫
log sin v − log v = 2 log x + log c
sin v
 sin v 
2
log 
= cx 2
 = log cx ⇒
v
 v 
x
y
 y
sin   = cx 2 ⇒ sin   = cxy
y
x
x
Non- Homogeneous Equations
Equations Reducible to Homogeneous Form:
The differential equation of the form
dy
ax + by + c
=
is called non homogeneous differential
dx a ′x + b′y + c′
equation.
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Very Short Answer Questions
I.
Solve the following differential equations.
1.
dy
12x + 5y − 9
=−
dx
5x + 2y − 4
Sol. from given d.e.
b = –5, a = 5 ⇒ b = –a
(5x + 2y – 4)dy = –(12x + 5y – 9)dx
(5x + 2y – 4)dy + (12x + 5y – 9)dx = 0
5(x dy+y dx) + 2y dy–4 dy+12x dx – 9 dx = 0
Integrating 5xy + y2 – 4y + 6x2 – 9x = c.
2.
dy −3x − 2y + 5
=
dx 2x + 3y + 5
Sol. From given d.e.
b = –2, a = 2 ⇒ b = –a
(2x + 3y + 5)dy = (–3x – 2y + 5)dx
2x dy + 3y dy + 5dy = –3x dx –2y dx + 5 dx
2x dy + 3y dy + 5dy + 3x dx –2y dx + 5 dx = 0
Integrating
2xy +
3 2 3 2
y + x + 5y − 5x = c
2
2
4xy + 3y 2 + 3x 2 − 10x + 10y = 2c = c′
Solution is
4xy + 3(x2 + y2) – 10(x – y) = c.
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3.
dy −3x − 2y + 5
=
dx
2x + 3y − 5
Sol.
dy −3x − 2y + 5
=
dx
2x + 3y − 5
Here b = –2, a′ = 2, b = –a′
(2x + 3y – 5)dy = (–3x – 2y + 5)dx
⇒ 2(x dy+y dx) + (3y – 5)dy + (3x – 5)dx = 0
⇒ 2d xy + (3y – 5)dy + (3x – 5)dx = 0
Now integrating term by term, we get
2 ∫ d(xy) + ∫ (3y − 5)dy + ∫ (3x − 5)dx = 0
y2
x2
c
⇒ 2xy + 3 − 5y + 3 − 5x =
2
2
2
(or)3x 2 + 4xy + 3y 2 − 10x − 10y = c
Which is the required solution.
4.
2(x − 3y + 1)
dy
= 4x − 2y + 1
dx
Sol. (2x – 6y + 2)dy = (4x – 2y + 1)dx
(2x – 6y + 2)dy – (4x – 2y + 1)dx = 0
2(x dy + y dx) – 6y dy + 2 dy – 4x dx – dx = 0
Integrating
2xy – 3y2 – 2x2 + 2y – x = c.
5.
dy x − y + 2
=
dx x + y − 1
Sol. b = –1, a′ = a ⇒ b = –a′
(x + y – 1)dy = (x – y + 2)dx
(x + y – 1)dy – (x – y + 2)dx = 0
(x dy + y dx) + y dy – dy – x dx – 2 dx = 0
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Integrating
xy +
y2 x 2
−
− y − 2x = c
2
2
2xy + y 2 − x 2 − 2y − 4x = 2c = c′
6.
dy 2x − y + 1
=
dx x + 2y − 3
Sol. b = –1, a′ = 1 ⇒ b = –a′
(x + 2y – 3)dy = (2x – y + 1)dx
(x + 2y – 3)dy – (2x – y + 1)dx = 0
(x dy + y dx) + 2y dy – 3 dy – 2x dx – dx = 0
Integrating:
xy + y2 – x2 – 3x – x = c
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Short Answer Questions
Solve the following differential equations.
1. (2x + 2y + 3)
Sol.
dy
=x+y+1
dx
( x + y) +1
dy
x + y +1
=
=
dx 2x + 2y + 3 2 ( x + y ) + 3
Let v = x + y so that
dv
dy
= 1+
dx
dx
dv
v + 1 2v + 3 + v + 1 3v + 4
= 1+
=
=
dx
2v + 3
2v + 3
2v + 3
2v + 3
dv = dx
3v + 4
2
1 3 ⋅ dv
dv + ∫
= dx
∫
3
9 3v + 4 ∫
2
1
v + log(3v + 4) = x + c
3
9
Multiplying with 9
6v + log(3v + 4) = 9x + 9c
6(x + y) + log[3(x + y) + 4] = 9x + c
i.e. log(3x + 3y + 4) = 3x – 6y + c
2.
dy 4x + 6y + 5
=
dx 2x + 3y + 4
Sol.
dy 4x + 6y + 5 2 ( 2x + 3y ) + 5
=
=
dx 2x + 3y + 4
2x + 3y + 4
Let v = 2x + 3y
dv
dy
dv
3(2v + 5)
= 2+3 ⇒
= 2+
dx
dx
dx
v+4
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2v + 8 + 6v + 15 8v + 23
=
v+4
v+4
v+4
dv = dx
8v + 23
=
1
9
dv
dv + ∫
= dx
∫
8
8 8v + 23 ∫
1
9
v + log(8v + 23) = x + c
8
64
Multiplying with 64
8v + 9 log(8v + 23) = 64x + 64c
8(2x + 3y) – 64x + 9log(16x + 24y + 23) = c′
Dividing with 8
9
2x + 3y − 8x + log(16x + 24y + 23) = c′′
8
9
3x − 6x + log(16x + 24y + 23) = c′′
8
Dividing with 3, solution is :
3
y − 2x + log(16x + 24y + 23) = k
8
3. (2x + y + 1)dx + (4x + 2y – 1)dy = 0
Sol.
dy
2x + y + 1
=−
dx
4x + 2y − 1
⇒ a1 = 2, b1 = 1, a 2 = 4, b 2 = 2
a1 2 1 b1
= = =
a 2 4 2 b2
Let 2x + y=v so that
dv
dy
= 2+
dx
dx
dv
v + 1 4v − 2 − v − 1 3(v − 1)
= 2−
=
=
dx
2v − 1
2v − 1
2v − 1
2v − 1
2v − 1
dv = dx ⇒
dv = 3dx
3(v − 1)
v −1
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
1 
∫  2 + v − 1  dv = 3∫ dx
2v + log(v − 1) = 3x + c
2v − 3x + log(v − 1) = c
2(2x + y) − 3x + log(2x + y − 1) = c
4x + 2y − 3x + log(2x + y − 1) = c
Solution is x + 2y + log(2x + y – 1) = c
4.
dy
2y + x + 1
=
dx 2x + 4y + 3
Sol.
dy
2y + x + 1
=
dx 2x + 4y + 3
Let v = x + 2y so that
dv
2dy
= 1+
dx
dx
dv
2(v + 1) 2v + 3 + 2v + 2 4v + 5
= 1+
=
=
dx
2v + 3
2v + 3
2v + 3
1

2v + 3
1
dv = dx ⇒ ∫  +
 dv = ∫ dx
4v + 5
 2 2(4v + 5) 
1
1 1
v + ⋅ log(4v + 5) = x + c
2
2 4
Multiplying with 8
4v + log(4v + 5) = 8x + 8c
4(x + 2y) − 8x + log[4(x + 2y) + 5] = c′
Solution is:
4x + 8y – 8x + log(4x + 8y + 5) = c′
8y – 4x + log(4x + 8y + 5) = c′
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5. (x + y – 1)dy = (x + y + 1)dx
Sol.
dy x + y + 1
=
dx x + y − 1
v=x+y⇒
dv
dy
= 1+
dx
dx
dv
v + 1 v − 1 + v + 1 2v
= 1+
=
=
dx
v −1
v −1
v −1
v −1
 1
dv = 2 ∫ dx ⇒ ∫ 1 −  dv = 2x + c
v
 v
v − log v = 2x + c
∫
x + y − log(x + y) = 2x − c
(x – y) + log (x + y) = c is the required solution.
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Long Answer Questions
Solve the following differential equations.
1.
dy 3y − 7x + 7
=
dx 3x − 7y − 3
Sol.
⇒ a1 = −7, b1 = 3, a 2 = 3, b 2 = −7
a1 −7 b1
3
=
,
=
a2
3 b 2 −7
∴
a1 b1
≠
a 2 b2
Let x = x + h, y = y + k
Where 3k − 7h + 7 = 0 and3h − 7k − 3 = 0 and
dy dY
=
dx dX
Solving these equations,
h = 0 and k = 1
dy 3y − 7x
=
which is a homogeneous d.e.
dx 3x − 7y
Put y = vx ⇒
v+x
x
dy
dv
= v+x
dx
dx
dv x(3v − 7)
=
dx x(3 − 7v)
dv 3v − 7
3v − 7 − 3v + 7v 2
=
−v=
dx 3 − 7v
3 − 7v
7v 2 − 7 7v 2 − 7
=
=
3 − 7v
3 − 7v
3 − 7v
7v 2 − 7
3
=
dx
x
7vdv
∫ 7v2 − 7 dv − ∫ 7v2 − 7 = ∫
ln x =
dx
x
3
v −1 1
ln
− ln | v 2 − 1| 14 log x–log c
14 v + 1 2
x = 3log
v −1
− 7 log | v 2 − 1|⇒ 14 ln x − ln c
v +1
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= 3ln(v − 1) − 3ln(v + 1) − 7 ln(v + 1) − 7 ln(v − 1)
14 ln x − ln c = −10 ln(v + 1) − 4 ln(v − 1)
ln(v + 1)5 + ln(v − 1) 2 + ln x 7 = ln c
(v + 1)5 ⋅ (v − 1)2 ⋅ x 7 = c
5
2
y  y  7
 + 1  − 1 x = c
x  x 
(y − x)2 (y + x)5 = c
[y − (x − 1)]2 (y + x − 1)5 = c
Solution is [y − x + 1)]2 (y + x − 1)5 = c
2.
dy
6x + 5y − 7
=
dx 2x + 18y − 14
Sol.
dy
6x + 5y − 7
=
dx 2x + 18y − 14
x = X + h, y = Y + k
dY dy
6(X + h) + 5(Y + k) − 7
=
=
dX dx 2(X + h) + 18(Y + k) − 14
=
(6X + 5Y) + (6h + 5k − 7)
(2X + 18Y) + (2h + 18k − 14)
h
+5
18
k
−7
−14
1
+6
+2
5
18
h
k
1
=
=
−70 + 126 −14 + 84 108 − 10
h=
56 4
70 5
= ,k =
=
98 7
98 7
dY 6X + 5Y
=
dX 2X + 18Y
dY
dV
Y = VX ⇒
= V+X
dX
dX
dV 6X + 5VX
X(6 + 5V)
V+X
=
=
dX 2X + 18VX X(2 + 18V)
X
dV 6 + 5V
6 + 5V − 2V − 18V 2
=
−V =
dX 2 + 18V
2 + 18V
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6 + 3V − 18V 2 3(2 + V − 6V 2 )
=
=
2 + 18V
2 + 18V
2 + 18V
∫ 6V 2 − V − 2 dV = −3∫
2 + 18V
Let
6V − V − 2
2
=
dX
X
A
B
+
3V − 2 2V + 1
Multiplying with (3V – 2)(2V + 1)
2 + 18V = A(2V + 1) + B(3V – 2)
2
4 
⇒ 2 + 12 = A  + 1
3
3 
7
14 = A − ⇒ A = 6
3
1
 3

V = − ⇒ 2 − 9 = B − − 2
2
 2

V=
7
−7 = − B ⇒ B = 2
2
2 
dX
 6
∫  3V − 2 + 2V + 1  dV = −3∫ X
2 log(3V – 2) + log(2V + 1) = –3 log X + log c
log (3V – 2)2 (2V + 1) + log X3 = log c
log X3 (3V – 2)2 (2V + 1) = log c
X3 (3V – 2)2 (2V + 1) = c
3  3Y
2
  2Y 
X 
− 2 
+ 1 = c
 X
  X

X3
(3Y − 2X)2 (2Y + X)
=c
X
X2
 
5 
4 
 3 y − 7  − 2  x − 7  
 

 
2
 
5 
4 
 2 y −  +  x −  = c
7 
7 
 
(3y − 2x − 1)2 (2y + x − 2)
=c
7
72
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Solution is:
(3y – 2x – 1)2 (x + 2y – 2) = 343c = c″
3.
dy 10x + 8y − 12
+
=0
dx 7x + 5y − 9
Sol.
dy 10x + 8y − 12
+
=0
dx 7x + 5y − 9
x = X + h, y = Y + k
⇒
dY dy
=
dX dx
dY 10(X + h) + 8(Y + k) − 12
+
=0
dX 7(X + h) + 5(Y + k) − 9
dY (10X + 8Y) + (10h + 8k − 12)
+
dX
(7X + 5Y) + (7h + 5k − 9)
Choose h and k so that:
10h + 8k – 12 = 0, 7h + 5k – 9 = 0
h
+8
+5
k
−12
−9
I
10
+7
8
5
h
k
1
=
=
−72 + 60 −84 + 90 50 − 56
h=
−12
6
= 2, k =
= −1
−6
−6
dY
10X + 8Y
=−
dX
7X + 5Y
dY
dV
= V+X
dX
dX
dV
10X + 8VX
X(10 + 8V)
V+X
=−
=−
dX
7X + 5VX
X(7 + 5V)
Y = VX ⇒
X
dV
10 + 8V
−10 − 8V − 7V − 5V 2
=−
−V =
dX
7 + 5V
7 + 5V
X
dV
5(V 2 + 3V + 2)
=−
dX
7 + 5V
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5V + 7
dX
X
5V + 7
A
B
=
+
(V + 1)(V + 2) V + 1 V + 2
∫ (V + 1)(V + 2) dV = −5∫
5V + 7 = A(V + 2) + B(V + 1)
V = −1 ⇒ 2 = A(−1 + 2) = A ⇒ A = 2V
V = −2 ⇒ −3 = B(−2 + 1) = − B, B = 3
 2
3

∫  V + 1 + V + 2  dV = −5∫
dX
X
2 log(V + 1) + 3 log (V + 2) = –5 log X + c
c = 2 log(V + 1) + 3 log(V + 2) + 5 log X = log(V + 1)2 (V + 2)3 x5
2
3
(Y + X)2 (Y + 2X)3 5
Y  Y

log  + 1  + 2  X5 = log
X
X  X

X2
X3
⇒ (Y + X) 2 (Y + 2X)3 = ec = c′
(Y + 1 − X − 2)2 (Y + 1 − 2x − 4)3 = c
Solution is : (x + y – 1)2 (2x + y – 3)3 = c
4. (x – y – 2)dx + (x – 2y – 3)dy = 0
dy − x + y + 2
=
dx x − 2y − 3
Sol. Given equation is
Let x = X + h, y = Y + k
dy −(X + h) + (Y + k) + 2
−X + y + (k − h + 2)
=
=
dx (X + h) − 2(Y + k) − 3 (X − 2Y) + (h − 2k − 3)
Choose h and k so that:
–h + k + 2 = 0, h – 2k – 3 = 0
h
−1
−2
k
−2
−3
I
1
1
−1
−2
h
k
1
=
=
+3 − 4 −2 + 3 −2 + 1
h = 1, k = –1
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dY − X + Y
=
dX X − 2Y
Put Y = VX so that
V+X
X
∫
dY
dV
=V+X
dX
dX
dV X(−1 + V) −1 + V
=
=
dX X(1 − 2V) 1 − 2V
dV −1 + V
−1 + V − V + 2V 2 2V 2 − 1
=
−V =
=
dX 1 − 2V
1 − 2V
1 − 2V
(1 − 2V)dV
2V − 1
2
dX
∫ X =∫
=
=∫
dX
X
1
− (−4V) − 1
2
dV
1 − 2V 2
1 (−4VdV)
dV
−∫
∫
2
2 1 − 2V
1 − 2V 2
1
1
dV
log | x |= − log |1 − 2V 2 | − ∫
2
2  1 2
2

 −V
 2
1
+V
1
1
1
2
2
= − log |1 − 2V | − ⋅
log
+ log c 2 log | x |= − log |1 − 2V 2 |
1
2
2 2
−V
2
1
+V
1
2
−
log
+ log C
1
 1 
−
V
2

2
 2
1
1+ V 2
log
+ log c
2
1− V 2
2 log | x | + log |1 − 2V 2 |= −
log X 2 (1 − 2V 2 ) = −
1
X+Y 2
log
+ log c
2
X−Y 2
1/ 2
 X−Y 2 
log | X − 2Y |= log c 

X+Y 2 
2
2
1/ 2
 X−Y 2 
∴ X − 2Y = c 
 X + Y 2 


2
2
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Substituting X = x–h = x – 1 , Y = y–k = y + 1
1/ 2
 x − 1 − (y + 1) 2 
(x − 1) − 2(y + 1) = c 
 x − 1 + (y + 1) 2 


2
2
1/ 2
 x − y 2 −1 − 2 
(x − 2y − 2x − 4y − 1) = c 
 x + y 2 − 1 + 2 


2
2
5. (x – y)dy = (x + y + 1) dx
Sol.
dy x + y + 1
=
dx
x−y
x = X + h, y = Y + k
dy X + h + Y + k + 1 (X + Y) + (h + k + 1)
=
=
dx (X + h) − (Y + k)
(X − Y) + (h − k)
Choose h and k so that h + k + 1 = 0, h – k = 0
1
2
Solving h = − , k = −
∴
dY X + Y
=
dX X − Y
Put Y = VX ⇒
V+X
X
∴
1
2
dY
dV
=V+X
dX
dX
dV X(1 + V)
=
dX X(1 − V)
dV 1 + V
1 + V − V + V2 1 + V2
=
−V =
=
dX 1 − V
1− V
1− V
(1 − V)dV
1+ V
dV
2
=
dX
X
1 2VdV
∫ 1 + V2 − 2 ∫ 1 + V2 = ∫
dX
X
1
tan −1 V − log(1 + V 2 ) = log x + log c
2
2 tan −1 V = log(1 + V) 2 + 2 log x + 2 log c
= log c 2 x 2 (1 + V 2 )
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 y2 
Y
2 tan −1   = log c 2 x 2 1 + 2 
 x 
X


1

y+ 

2 = log c 2 (Y 2 + X 2 )
2 tan −1 
1
x+ 
2

2
2

1 
1 
1
 2y + 1 
2 
2 2
2
2 tan −1 
=
log
c
x
+
+
y
+



 
  = log c  x + y + x + y + 
2 
2  
2
 2x + 1 


6. (2x + 3y – 8)dx = (x + y – 3)dy
Sol.
dy 2x + 3y − 8
=
dx
x + y −3
x = X + h, y = Y + k so that
dY dy
=
dX dx
dY 2(X + h) + 3(Y + k) − 8
=
dX
(X + h) + (Y + k) − 3
(2X + 3Y) + (2h + 3k − 8)
=
(X + Y) + (h + k − 3)
Choose h and k so that:
2h + 3k – 8 = 0, h + k – 3 = 0
h
k
−8
−3
3
1
I
2
1
3
1
h
k
1
=
=
−9 + 8 −8 + 6 2 − 3
h = 1, k = 2
dY 2X + 3Y
∴
=
dX
X+Y
Put Y = VX so that
V+X
=
dY
dV
=V+X
dX
dX
dV X(2 + 3V)
dV 2 + 3V
=
⇒X
=
−V
dX
X(1 + V)
dX 1 + V
2 + 3V − V − V 2 2 + 2V − V 2
=
1+ V
1+ V
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(1 + V)dV
∫ 2 + 2V − V 2 = ∫
Consider
dX
X
(1 + V)dV
∫ 2 + 2V − V 2
Let (1 + V) = A(2 – 2V) + B
Equating the coefficients of V
1 = –2A ⇒ A = –1/2
Equating constants:
1 = 2A + B = –1 + B ⇒ B = 2
(1 + V)dV
1 (2 − 2V)dV
dV
∫ 2 + 2V − V 2 = − 2 ∫ 2 + 2V − V 2 + 2∫ 2 + 2V − V 2
1
dV
= − log(2 + 2V − V 2 ) + 2 ∫
2
( 3) 2 − (V − 1)2
1
1
3 + V −1
= − log(2 + 2V − V 2 ) + 2
log
2
2 3
3 − V +1
1
1
V + ( 3 − 1)
= − log(2 + 2V − V 2 ) +
log
2
3
− V + ( 3 + 1)
Y
+ ( 3 − 1)

1
2Y Y 2  1
X
= − log  2 +
−
log
+

Y
2
X X 2 
3

− + 3 +1
X
1
1
Y + ( 3 − 1)X
= − log(2X 2 + 2XY − Y 2 ) +
log
2
3
Y − ( 3 + 1)Y
1
1
[Y + 3 − 1)X]
∴ log X + c = − log(2X 2 + 2XY − Y 2 ) +
log
2
3
Y − ( 3 + 1)X
7.
dy x + 2y + 3
=
dx 2x + 3y + 4
Sol. Let x = X + h, y = Y + k so that
dY dy
=
dX dx
dY (X + h) + 2(Y + k) + 3
=
dX 2(X + h) + 3(Y + k) + 4
(X + 2Y) + (h + 2k + 3)
=
(2X + 3Y) + (2h + 3k + 4)
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Choose h and k so that:
h + 2k + 3 = 0, 2h + 3k + 4 = 0
h
k
3
4
2
3
I
1
2
2
3
h
k
1
=
=
8−9 6− 4 3− 4
2
−1
h=
= 1, k =
= −2
−1
−1
dY X + 2Y
=
dX 2X + 3Y
This is a homogeneous equation
dY
dV
= V+X
dX
dX
dV X(1 + 2V)
V+X
=
dX X(2 + 3V)
Y = VX ⇒
X
dV 1 + 2V
1 + 2V − 2V − 3V 2
=
−V =
dX 2 + 3V
2 + 3V
(2 − 3V)dV
dX
X
1 − 3V
dV
1 −6VdV
dX
− ∫
=∫
2∫
2
2
2 1 − 3V
X
1 − 3V
2
=
1
+V
2
dV
1
2 1
1
3
2
− log |1 − 3V | = log X + log c
log
− log |1 − 3V 2 |= log cx
∫
2
1
1
3  1 
2
3 2⋅
2
−V 2
−
V


3
3
 3
1
3 3
log
1 + 3V
log |1 − 3V 2 |= log cy
1 − 3V
3Y
1
1
3y 2
X
log
− log 1 − 2 = log cy
3 3
3Y 2
x
1−
X
1+
1
X + 3Y 1
X 2 − 3Y 2
log
− log
= log cx
3
X − 3Y 2
X2
Where X = x – 1, Y = y + 2
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1 
log(X + 3Y) − log(X − 3) 
3
1
− log(X + 3Y) + log(X − 3Y) − 2 log X
2
= log CX
(
)
1 
 1
log(X + 3Y) 
−
 − log(X − 3Y)
2
 3
1 1 
2+
 + log X = log c + log X
3

2− 3
2+ 3
log(X + 3Y)
log(X − 3Y)
2 3
2 3
= log c
(2 − 3)
(2 + 3)
log(X + 3Y)
2
2
log(X − 3Y) = 3c′ where c′ = log c
i.e.
8.
dy 2x + 9y − 20
=
dx 6x + 2y − 10
Sol. Given equation is
dy 2x + 9y − 20
=
dx 6x + 2y − 10
Let x = X + h, y = Y + k so that
dY dy
=
dX dx
dY 2(X + h) + 9(Y + k) − 20
=
dX 6(X + h) + 2(Y + k) − 10
(2X + 9Y) + (2h + 9k − 20)
=
(6X + 2Y) + (6h + 2k − 10)
Choose h and k so that:
2h + 9k – 20 = 0, 6h + 2k – 10 = 0
h
9
2
k
−20
−10
I
2
6
9
2
h
k
1
=
=
−90 + 40 −120 + 20 4 − 54
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−50
−100
= 1, k =
=2
−50
−50
dY 2X + 9Y
∴
=
dX 6X + 2Y
h=
This is homogeneous equation
dY
dV
=V+X
dX
dX
dV x(2 + 9V)
V+X
=
dX x(6 + 2V)
y = VX ⇒
X
dV 2 + 9V
2 + 9V − 6V − 2V 2
=
−V =
dX 6 + 2V
6 + 2V
6 + 2V ⋅ dY
2 + 3V − 2V
2
=
dX
X
(6 + 2V)dV
∫ (1 + 2V)(2 − V) = ∫
Let
dx
x
… (1)
6 + 2V
A
B
=
+
(1 + 2V)(2 − V) 1 + 2V 2 − V
6 + 2V = A(2 – V) + B(1 + 2V)
V = 2 ⇒ 10 = B(5) ⇒ B = 2
1
5
V = − ⇒ 5 = A  ⇒ A = 2
2
2
6 + 2V
2
2
=
+
(1 + 2V)(2 − V) 1 + 2V 2 − V
(6 + 2V)dV
2dV
dV
∫ (1 + 2V)(2 − V) = ∫ 1 + 2V − 2∫ V − 2
= log(1 + 2V) − 2 log(V − 2)
From (1) we get
log(1 + 2V) − log(V − 2)2 = log X − log c
1 + 2V
X
1 + 2V
X
log
= log ⇒
=
2
2
c
c
(V − 2)
(V − 2)
X
2V x  Y

1 + 2V = (V − 2)2 ⇒ 1 +
=  − 2
c
X cX

2
X + 2Y x (Y − 2X)2
=
X
c
X2
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Solution is (X + 2Y) = (Y – 2X)2
Where X = x – 1, Y = y – 2
c(x – 1 + 2y – 4) = (y – 2 – 2x + 2)2
c(x + 2y – 5) = (y – 2x)2 = (2x – y)2
Linear Differential Equations
Linear Equations: A differential equation of the form
dy
+ Py = Q , where P and Q are functions of
dx
x only is called a linear differential equation of the first order in y.
Bernoulli’s Equation: An equation of the form
dy
+ Py = Qy n , where P and Q are functions of x
dx
only, is called a Bernoulli’s equation
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Very Short Answer Questions
I.
Find the I.F. of the following differential equations by transforming them into linear form.
1.
x
dy
− y = 2x 2 sec2 2x
dx
Sol. x
dy
− y = 2x 2 sec2 2x
dx
dx 1
− y = 2x sec2 2x which is linear in y .
dy x
I.F. = e ∫
2.
y
Sol. y
pdx
1
1
= ∫ − log x = e− log x = elog(1/x) =
x
x
e
dx
− x = 2y3
dy
dx
dx 1
− x = 2y3 ⇒
− x = 2y 2 which is linear equation in x.
dy
dy y
1
I.F. = e
∫ pdy
=e
∫ − y dy
= e− log y = elog(1/ y) =
1
y
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Short Answer Questions
Solve the following differential equations.
1.
dy
+ y tan x = cos3 x
dx
Sol.
dy
+ y tan x = cos3 x which is linear d.eq. in y.
dx
I.F. = e ∫
p dx
= e∫
tan x dx
= elog(sec x) = sec x
Solution of the equation is
y. I.F. = y.I.F = ∫ Q. I.F. dx
⇒ y sec x = ∫ sec x cos3 x dx = ∫ cos 2 x dx
=
1
(1 + cos 2x)dx =
2∫
1
sin 2x 
x+
+c
2
2 
2y
= x + sin x ⋅ cos x + c
cos x
Solution is:
2y = x cos x + sin x ⋅ cos2 x + c ⋅ cos x
2.
dy
+ y sec x = tan x
dx
Sol.
dy
+ y sec x = tan x which is l.d.e. in y
dx
I.F. = e ∫
Sol is
sec xdx
= elog(sec x + tan x) = sec x + tan x
y.I.F. = y.I.F = ∫ Q. I.F. dx
y(sec x + tan x) = ∫ tan x(sec x + tan x)dx
= ∫ (sec x ⋅ tan x + tan 2 x)dx
= ∫ (sec x ⋅ tan x + sec2 x − 1)dx
Solution is
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y(sec x + tan x) = sec x + tan x − x + c
3.
dy
− y tan x = e x sec x Which is l. d.e. in y.
dx
− tan x dx
Sol. I.F. = e ∫
= elog cos x = cos x
Sol is
y.I.F. = y.I.F = ∫ Q. I.F. dx
y cos x = ∫ e x sec x cos xdx = ∫ e x dx = e x + c
4.
x
dy
+ 2y = log x
dx
2
Sol. I.F. = e
∫ x dx
2
= e2log x = elog x = x 2
Solution is:
∴ y ⋅ x2 = ∫ x2
log x
dx = ∫ x log xdx
x
 x2  1
x2
x2
1
x2
1
log x −
+c
= log x   − ∫ x 2 dx =
log x − ∫ x dx =
 2  2
2
4
x
2
2
 
5.
(1 + x 2 )
−1
dy
+ y = e tan x
dx
−1
dy
1
e tan x
Sol. +
⋅
=
which linear differential equation in y.
y
dx 1 + x 2
1+ x2
I.F. = e
Sol is
y⋅e
∫ pdx
dx
=e
∫ 1+ x 2
= e tan
−1
x
y.I.F. = y.I.F = ∫ Q. I.F. dx
tan −1 x
Consider
=∫
∫
(e tan
−1
x 2
)
1+ x
(e tan
−1
2
dx
x 2
1+ x2
)
dx
… (1)
put tan −1 x = t ⇒
dx
1+ x2
= dt
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e 2t e2 tan
= ∫ (e ) dt = ∫ e dt =
=
2
2
t 2
−1
2t
Solution is y ⋅ e
2y ⋅ e tan
−1
x
tan −1 x
= e 2 tan
e2 tan
=
2
−1
x
−1
x
+
x
c
2
+c
6.
dy 2y
+
= 2x 2
dx x
Sol.
dy 2y
+
= 2x 2 which is l.d.e. in y.
dx x
2
I.F. = e
∫ x dx
Sol is
2
= e2log x = elog x = x 2
y.I.F. = y.I.F = ∫ Q. I.F. dx
y ⋅ x 2 = ∫ 2x 4dx =
7.
2x 5
+c
5
dy
4x
1
+
y=
2
dx 1 + x
(1 + x 2 ) 2
4x
Sol. I.F. = e
∫ 1+ x 2 dx
= e2log(1+ x
= elog(1+ x
2 2
)
2
)
= (1 + x 2 )2
∴ Solution is y(1 + x 2 ) 2 = ∫ dx = x + c
dy
+ y = (1 + x)e x
dx
8.
x
Sol.
dy 1
(1 + x)e x
+ ⋅y =
dx x
x
1
I.F. = e
∫ x dx
= elog x = x
y ⋅ x = ∫ (1 + x)e x dx = x ⋅ e x + c
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9.
dy 3x 2
1+ x2
+
y=
dx 1 + x 3
1 + x3
Sol.
dy 3x 2
1+ x2
+
y
=
which is linear differential equation in y .
dx 1 + x 3
1 + x3
3x 2
I.F. = e
Sol is
∫ 1+ x 3 dx
= elog(1+ x ) = 1 + x 3
3
y.I.F. = y.I.F = ∫ Q. I.F. dx
x3
+c
3
y(1 + x 3 ) = ∫ (1 + x 2 )dx = x +
10.
dy
− y = −2e− x
dx
Sol. I.F. = e ∫
− dx
= e− x
y ⋅ e− x = −2∫ e −2x dx = e−2x + c
y = e − x + ce x
11. (1 + x 2 )
Sol.
dy
+ y = tan −1 x .
dx
dy
1
tan −1 x
+
⋅
y
=
which is linear differential equation in y.
dx 1 + x 2
1+ x2
dx
I.F. e
Sol is
y⋅e
∫ 1+ x 2
= e tan
−1
x
y.I.F. = y.I.F = ∫ Q. I.F. dx
tan −1 x
= ∫ tan
−1
x
e tan
−1
x
1+ x2
Put t = tan–1x so that dt =
dx
dx
1+ x2
R.H.S. = ∫ t ⋅ e t dt = t ⋅ e t − ∫ e t dt = t ⋅ e t − e t
Solution is: y.e tan
−1
x
= e tan
−1
x
(tan −1 x − 1) + c
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y = tan −1 x − 1 + c ⋅ e − tan
12.
−1
x
dy
+ y tan x = sin x .
dx
Sol. I.F. = e ∫
tan x dx
= elog sec x = sec x
y sec x = ∫ sin x ⋅ sec x dx
= ∫ tan x dx = log sec x + c
Long Answer Questions
Solve the following differential equations.
dy
+ y sin x = sec 2 x
dx
1.
cos x
Sol.
dy
+ tan x ⋅ y = sec3 x which is l.d.e in y
dx
I.F. = e ∫
Sol is
tan x dx
= elog sec x = sec x
y.I.F. = y.I.F = ∫ Q. I.F. dx
y ⋅ sec x = ∫ sec 4 xdx = ∫ (1 + tan 2 x) sec2 xdx
= tan x +
tan 3 x
+c
3
2.
sec x ⋅ dy = (y + sin x)dx
Sol.
dy y + sin x
=
= y cos x + sin x ⋅ cos x
dx
sec x
dy
− y cos x = sin x ⋅ cos x which is l..d.e in y
dx
− cos xdx
I.F. = e ∫
= e − sin x
Sol is
y.I.F. = y.I.F = ∫ Q. I.F. dx
y ⋅ e− sin x = ∫ e− sin x ⋅ sin x ⋅ cos x ⋅ dx
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Consider ∫ e − sin x ⋅ sin x ⋅ cos x ⋅ dx
t = – sin x ⇒ dt = –cos x dx
∫e
− sin x
⋅ sin x ⋅ cos x dx = + ∫ e t ⋅ t dt
= t ⋅ e t − e t + c = e − sin x (− sin x − 1) + c
y ⋅ e− sin x = −e− sin x (sin x + 1) + c
or y = −(sin x + 1) + c ⋅ esin x
dy
+ y = 2 log x
dx
3.
x log x ⋅
Sol.
dy
1
2
+
y=
dx x log x
x
dx
I.F. = e
∫ x log x
y log x = 2 ∫
= elog(log x) = log x
log x
dx = (log x)2 + c
x
(x + y + 1)
dy
=1
dx
Sol. (x + y + 1)
dy
=1
dx
4.
dx
dx
= x + y +1 ⇒
− x = y + 1 which is l.d.e in x.
dy
dy
I.F. = e ∫
pdy
= e∫
− dy
= e− y
sol is x.I.F.= ∫ Q.IFdy
x ⋅ e − y = ∫ e − y (y + 1)dy = −(y + 1)e− y + ∫ e− y dy
= −(y + 1)e− y − e− y = −(y + 2)e− y + c
x = −(y + 2) + c ⋅ e y
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dy
− y = x 3 (x − 1)3 .
dx
5. Solve x(x − 1)
Sol.
dy
1
−
y = x 2 (x − 1) 2 which is l.d.e in y
dx x(x − 1)
1
dx
I.F. = e
=e
=e
∫ − x(x −1)
log x −log(x −1)
Sol is
y⋅
∫ pdx
=e
log
=e
x
x −1
1 
∫  x − x −1 
=
dx
x
x −1
y.I.F. = y.I.F = ∫ Q. I.F. dx
x
x
= ∫ x 2 (x − 1)2
dx = ∫ x 3 (x − 1)dx
x −1
(x − 1)
Hence solution is
xy
x5 x 4
=
−
+c
x −1 5
4
dy
=y
dx
6.
(x + 2y3 )
Sol.
dy x + 2y3 x
=
= + 2y 2
dx
y
y
dx 1
− x = 2y 2
dy y
1
I.F. = e
x⋅
∫ − y dy
= e − log y = elog1/ y =
1
y
1
= 2ydy = y 2 + c
y ∫
Solution is: x = y(y 2 + c)
7. Solve (1 − x 2 )
Sol.
dy
+ 2xy = x 1 − x 2 .
dx
dy
2x
x
+
y
=
dx 1 − x 2
1 − x2
2x
I.F. = e
∫ 1− x 2 dx
= e − log(1− x
2
)
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= elog(1− x
y
1− x
2
2 −1
)
=
1
1− x2
xdx
=∫
(1 − x )
2 3/ 2
= (1 − x 2 ) −1/ 2 + c
(or)y = 1 − x 2 + c(1 − x 2 )
dy
− (x − 2)y = x 3 (2x − 1)
dx
8.
x(x − 1)
Sol.
dy
x−2
x 3 (2x − 1)
−
y=
dx x(x − 1)
x(x − 1)
2− x
∫ x(x −1)
2−x
A
B
= +
x(x − 1) x x − 1
2 − x = A(x − 1) + Bx
x = 0 ⇒ 2 = − A ⇒ A = −2
x =1⇒1= B ⇒ B =1
I.F. = e
dx ⇒
2−x
−2
1
=
+
x(x − 1) x x + 1
2−x
dx
dx
∫ x(x − 1) dx = −2∫ x + ∫ x − 1
= − log x + log(x − 1) = log
I.F. = e
y
x −1
x2
log
x −1
x2
=∫
=
x −1
x2
x −1
x2
x 3 (2x − 1) x − 1
⋅
dx
x(x − 1) x 2
= ∫ (2x − 1)dx = x 2 − x + c
Solution is y(x – 1) = x2(x2 – x + c)
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9.
dy 2 3
(x y + xy) = 1
dx
Sol.
dy 2 3
(x y + xy) = 1
dx
dx
= xy + x 2 y3
dy
⇒
dx
− xy = x 2 y3 ---- (1)
dy
Which is Bernoulli’s equation
Dividing with x2,
1 dx 1
− y = y3
2 dy x
x
1
x
Put z = − so that
⇒
dz 1 dx
=
dy x 2 dy
dz
+ z ⋅ y = y3 ----2)
dy
Which is linear diff.eq. in z
I.F. = e ∫
ydy
= ey
2
/2
Sol is z.I.F = ∫ Q. I.F. dy
z ⋅ ey
2
/2
= ∫ y 3e y
2
/2
⋅ dy
put
= ∫ t ⋅ dt ⋅ e = e (t − 1) = e
t
z ⋅ ey
z=
2
/2
= ey
t
2
y2
= t ⇒ y dy = dt
2
y 2 /2 
y2 
− 1


 2

/2 
y2 
− 1 + c


2


2
2
y2
1 y2
− 1 + c ⋅ e− y /2 ⇒ − =
− 1 + c ⋅ e− y /2
2
x 2
 y2

2
−1 = x 
− 1 + c ⋅ e − y /2 
 2



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 y2
Hence solution is 1 + x 
 2
10.
Sol.
− 1 + c ⋅ e− y
2
/2 
 = 0

dy
+ x ⋅ sin 2y = x 3 cos 2 y
dx
1
dy 2sin y cos y
+
x = x3
2
dx
cos y
cos y
2
sec2 y
dy
+ 2 tan y.x = x 3
dx
z = tan y ⇒
dz
dy
= sec 2 y
dx
dx
dz
+ 2zx = x 3
dx
I.F. = e ∫
2xdx
2
= ex
2
2
z ⋅ e x = ∫ x 3 ⋅ e x dx
...(1)
2
Consider ∫ x 3 ⋅ e x dx
t = x 2 ⇒ dt = 2x ⋅ dx
2
2
3 x
2 x
∫ x ⋅ e dx = ∫ x ⋅ x ⋅ e dx
=
1
1
t ⋅ e t dt = e t (t − 1)
∫
2
2
2
= z ⋅ ex =
z=
1 x2 2
e (x − 1) + c
2
2
x2 −1
+ c ⋅ e− x
2
tan y =
2
x 2 −1
+ c ⋅ e− x
2
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
1  dy
11. y 2 +  x −  = 0
y  dx

Sol.

1  dy
y2 +  x − 
=0
y  dx


1  dy
= − y2
x− 
y  dx

dx x − 1/ y
x
1
=
=− 2 + 3
2
dy
−y
y
y
dx 1
1
+ 2 ⋅ x = 3 Which is l.d.e in x
dy y
y
1
I.F. = e
∫ y2 dy
= e−1/y
Sol is x.I.F = ∫ Q. I.F. dy
x ⋅ e −1/ y = ∫
e −1/ y
y3
dy
put −
… (1)
1
1
= z ⇒ 2 dy = dz
y
y
= ∫ z ⋅ ez dz = ez (z − 1)
 1 
x ⋅ e −1/y = −e−1/y  − − 1  + c
 y 
x
1/ y
e
=
1+ y
y ⋅ e1/y
+c
Hence solution is xy = 1 + y + cy e1/y.
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Problems for Practice
1 . Form the differential equation corresponding to the family of circles passing through the
origin and having centers on Y-axis.
Ans. (x 2 − y 2 )
2. Solve
dy
− 2xy = 0
dx
dy
= sin(x + y) + cos(x + y)
dx


Ans. x = log 1 + tan
x+y
+c
2 
y
x
3. Give the solution of x sin 2 dx = ydx − xdy which passes through the point (1, π/4).
Ans. cot
y
= log x + 1
x
(
4. Solve 2x − 10y3
dy
+y=0
) dx
2
Ans. I.F. = e
∫ y dy
(
2
= e2log y = elog y = y 2
5. Solve 1 + x 2
dy
+ 2xy − 4x
) dx
Ans. y(1 + x 2 ) = ∫ 4x 2 dx =
6. Solve
2
=0
4x 3
+c
3
2
1 dy
⋅ + y ⋅ e x = e(1− x)e .
x dx
Ans. 2y ⋅ e(x −1)e = x 2 + 2c
x
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7. Solve sin 2 x ⋅
dy
+ y = cot x .
dx
Ans. y ⋅ e− cot x = (cot x + 1)e− cot x + c
8. Find the equation of the equation x(x − 2)
dy
− 2(x − 1)y = x 3 (x − 2) which satisfies the
dx
condition that y = 9 when x = 3.
Ans.
y
= x + 2 log(x − 2)
x(x − 2)
9. Solve (1 + y 2 )dx = (tan −1 y − x)dy .
Ans. x ⋅ e tan
−1
y
= e tan
−1
y
(tan −1 y − 1) + c .
10. Solve (x 3 − 3xy 2 )dx + (3x 2 y − y3 )dy = 0 .
y2
Ans.
x2
y2
x2
−1
= cx
+1
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