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1
1&2
EN0101:Engineering mathematics 1
Module: IV
MODULE IV
ORDINARY DIFFERENTIAL EQUATIONS
DIFFERENTIAL EQUATION
An equation which involves differential coefficients is called a
differential equation
eg:
1+x2
1-y2
1. dy =
dx
2. d2 y + 2 dy -8y = 0
dx2
dx
3.
1+ (dy)2
dx
3/2
=
k d2 y
dx2
ORDER AND DEGREE OF A DIFFERENTIAL EQUATION
The order of a differential equation is the order of the highest
differential coefficient present in the differential equation .
The degree of a differential equation is the degree of the highest
derivative.
eg:
1. L d2q + R dq + q = Esinwt
dt2
dt c
2. cosx d2 y + sinx (dy)2 + 8y = tanx
dx2
dx
3.
[ 1+ (dy) ] 3
2
dx
=
(d2 y )2
dx2
2
The order of the above equations is 2. The degree of the equations 1 and 2 is 1.
Degree of equation 3 is 2.
LINEAR DIFFERENTIAL EQUATION
A linear differential equation is the differential equation in which the
dependent variable and its derivatives occur only in the first degree and are not multiplied
together.
The general linear differential equation of the nth order is of the form
dny + P1 dn-1y + P2 dn-2y + …………………….. + Pn-1 dy + Pn y = X
dxn
dxn-1
dxn-2
dx
where P1, P2…………………Pn-1, Pn and X are functions of x only.
A linear differential equation with constant coefficients is of the form
dny + a1 dn-1y + a2 dn-2y + …………………….. + an-1 dy + an y = X
dxn
dxn-1
dxn-2
dx
------------------------------( 1 )
Where a1, a2…………………an-1, an are constants and X is either a constant or a function
of x only.
THE OPERATOR
D
d ≡D
dx
d2 ≡ D2 …………………dn ≡ Dn.
dx2
dxn
Thus the symbol D is a differential operator or simply an operator . Thus the symbolic
form of equation ( 1 ) becomes
Dn y + a1 Dn-1 y + a2 Dn-2 y+ …………………….. + an-1 Dy + an y = X
( Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an ) y = X
ie f(D)y = X
where f(D) = Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an which is a
polynomial in D.
ie f(D) is a polynomial in D.
3
NOTE
The polynomial D can be treated as an algebraic quantity
D(u + v) = Du + Dv
D( λu ) = λ Du
DpDqu = Dp+q u
DpDqu = Dq Dp u
AUXILIARY EQUATION
Consider the differential equation
( Dn + P1 Dn-1 + P2 Dn-2 + …………………….. + Pn-1 D + Pn ) y = X
then“ the equation obtained by equating to zero the symbolic coefficient of y is called the
auxiliary equation ” briefly written as AE. Thus the AE of the above equation is,
Dn + P1 Dn-1 + P2 Dn-2 + …………………….. + Pn-1 D + Pn = 0
SOLUTION OF A DIFFERENTIAL EQUATION
An equation containing dependent variable y and independent variable x and
free from derivatives which satisfies the differential equation is called the solution of the
differential equation.
THEOREM
If y = u is the complete solution of the equation f(D) y = 0 called the
complementary function and y = v is the particular solution of the equation f(D)y =X
called particular integral of the equation then the complete solution of the equation
f(D)y = X is y = u + v.
ie complete solution is y = CF + PI.
STEPS FOR FINDING THE CF OF THE EQUATION
Consider the equation
dny + a1 dn-1y + a2 dn-2y + …………………….. + an-1 dy + an y = X
dxn
dxn-1
dxn-2
dx
1. Write the homogeneous equation in the symbolic form
ie (Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an )y = 0
4
2. Write down the auxiliary equation
ie Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an = 0 and solve
for D
3. Write the solution as follows
Roots for AE
1. m1 ,m2 ,m3 ,……….
Soln. CF
1. y = c1 em1x + c2 em2x +………
(Real and distinct roots)
2. m1 ,m1 ,m3 ,……….
2. y = (c1 + c2x) em1x + c3 em3x +………
(2 real and equal roots)
3. m1 ,m1 ,m1 ,……….
3.y = (c1+ c2x + c3 x2) em1x + c4 em4x
+……….
(3 real and equal roots)
4. α +iβ , α - iβ ,………
4. y = eαx ( c1 cosβx +c2 sinβx) +……
(A pair of imaginary
roots) .
5. α +iβ, α +iβ, α – iβ,
α - iβ ,……….
(Two pair of imaginary
roots be equal)
EXAMPLES
1. Solve d 3y – 7dy -6y = 0
dx 3 dx
Soln:
Symbolic form of the equation is
( D3- 7D -6) y = 0
Auxiliary equation is
5. y = eαx ( (c1+ c2 x)cosβx + (c3 + c4 x)
sinβx) +……
5
D3- 7D -6 = 0
Roots of auxiliary equation are
D = -1, -2, 3
CF is , y = C1e-x + C2e-2x +C3 e3x
2. Solve ( D3- 4D2 +4D) y = 0
Soln:
Auxiliary equation is
D3- 4D2 +4D = 0
D(D2 -4D+4) =0
D(D-2)2 =0
Roots of auxiliary equation are
D = 0, 2 , 2
CF is , y = (C1x + C2)e2x +C3 e0x
y = (C1x + C2)e2x +C3
3. Solve d 4y + 13d2y +36y = 0
dx 4
dx2
Soln:
Symbolic form of the equation is
( D4+ 13D2 +36) y = 0
Auxiliary equation is
D4+ 13D2 +36 = 0
ie (D2 +9) (D2 + 4) = 0
Roots of auxiliary equation are
D = ±3i , ±2i
CF is , y = e0x ( C1 cos3x + C2 sin3x ) + e0x ( C3 cos2x + C4 sin2x )
y = C1 cos3x + C2 sin3x + C3 cos2x + C4 sin2x
PROBLEMS
1. Solve d 4y + 4y = 0
dx 4
2. Solve (D 2+1)3 ( D2+D+1) 2 y = 0
3. Solve (D 2+1)2 ( D2+D+1) y = 0
THE INVERSE OPERATOR 1
f(D)
Definition : 1 X is that function of x ,free from arbitrary constants, which when
f(D
operated upon by f(D) gives X.
6
{(
)}
Thus f(D)
1/f(D) X = X
f(D) and 1/f(D) are inverse operators.
Theorem 1. (1/f(D))X is the particular integral of f(D)y = X
Theorem 2. (1/D) X = ∫ X dx
Theorem 3. ( 1/D-a ) X = eax ∫ Xe-ax dx
RULES FOR FINDING THE PARTICULAR INTEGRAL
Consider the differential equation ,
( Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an ) y = X
It can be written as f(D) y = X
P.I = 1 X
f(D)
Case 1
When X= eax
P.I = 1 eax
f(D)
= 1 eax
Replace D by a ,provided f(a) ≠ 0
f(a)
If f(a) = 0 , 1 eax = x 1 eax ,provided f 1(a) ≠ 0
f(D)
f 1(a)
If f 1(a) = 0 , 1 eax = x2 1 eax ,provided f 11(a) ≠ 0 and so on .
f(D)
f 11(a)
Case 11
When X= sin (ax+b) or cos(ax+b)
P.I = 1 sin (ax+b)
f(D2)
= 1 sin (ax+b)
Replace D2 by –a2 ,provided f(-a2) ≠ 0
2
f(-a )
and 1 cos (ax+b) = 1 cos (ax+b)
Replace D2 by –a2 ,provided f(-a2) ≠ 0
2
2
f(D )
f(-a )
If f(-a2) = 0 ,
and
and 1 cos (ax+b) = x 1 cos (ax+b)
f(D2)
f 1(-a2)
1 sin (ax+b) = x 1 sin (ax+b) ,provided f 1(-a2) ≠ 0
f(D2)
f 1(-a2)
If f 1(-a2) =0 , 1 cos (ax+b) = x2 1 cos (ax+b)
f(D2)
f 11(-a2)
7
1 sin (ax+b) = x2 1 sin (ax+b)
f(D2)
f11(-a2)
and
,
provided f 11(-a2) ≠ 0 and so on .
Case 111
When X = xm , m being a positive integer
P.I = 1 xm
f(D)
= ( f(D))-1 xm
Expand ( f(D))-1 in ascending powers of D as far as the term in Dm and operate on
xm term by term . Since the (m+ 1)th and higher derivatives of xm are zero, we need not
consider terms beyond Dm .
Case 1V
When X= eax V , V being a function of x.
P.I = 1 eax V
f(D)
= eax 1 V
f(D+a)
Case V
When X is any other function of x
Let f(D) = (D-m1 ) (D-m2) (D-m3)…………… (D-mn )
Then P.I = 1 X
f(D)
=
1
X
(D-m1 ) (D-m2) (D-m3)…………… (D-mn )
= A1 + A2 +……………………………………+ An
X
D-m1 D-m2
D-mn
=
A1 X
D-m1
+
A2 X
D-m2
+……………………………………+
= A1 em1x ∫ X e - m1x dx + A2 em2x ∫ X e – m2x dx
EXAMPLES
1.
Solve d2y + 4dy +4y = e2x – e-2x
dx2
dx
Soln:
A.E is m2 + 4m +4 =0
(m +2) 2 = 0
m = -2 ,-2
An X
D-mn
+…………+
An emnx ∫ X e - mnx dx
8
C.F is y = ( c1 +c2 x)e-2x
1 e2x – e-2x
(D+2)2
= 1 e2x – 1 e-2x
(D+2)2
(D+2)2
P.I =
= e2x – xe-2x
16 2(D+2)
= e2x – x2e-2x
16
2
Hence the C.S is ,
y = ( c1 +c2 x)e-2x + e2x – x2e-2x
16
2
2. Solve d3y + 4dy + = sin2x
dx3
dx
Soln:
A.E is m3 + 4m =0
m( m2 +4) = 0
m = 0 , ±2i
C.F is y = c1 +c2 cos2x + c3 sin2x
P.I =
1 sin2x
D(D2 +4)
= x
1 sin2x
3D2 +4
= x
1 sin2x
3. -4 +4
= x
1 sin2x
-8
Hence the C.S is ,
y = c1 +c2 cos2x + c3 sin2x - x sin2x
8
3. Solve 2 d2y + 5dy +2y = 5+ 2x
dx2
dx
Soln:
9
A.E is 2m2 + 5m +2 =0
m = -2 ,- 1/2
C.F is y = c1 e-2x + c2 e-x/2
P.I =
1 (5+ 2x)
2D + 5D +2
= 1 ( 1+D2 + 5D )-1 (5+ 2x)
2
2
2
= 1 ( 1- (D2 + 5D) + …………. ) (5+ 2x)
2
2
= 1 ( 1- 5D ) (5+ 2x)
2
2
= 1 (5+ 2x - 5.2 )
2
2
= 1 (2x)
2
= x
Hence the C.S is ,
y = c1 e-2x + c2 e-x/2 + x
4. Find the P I of (D2 -2D +4)y = ex cosx
Soln:
PI = 1X
f(D)
=
1 ex cosx
2
D -2D +4
= ex 1 cosx
(D +1) 2 -2( D +1) +4
= ex 1 cosx
D2 +3
=
ex 1 cosx
-1+3
= ex cosx
2
5. Solve d2y + y = cosec x
dx2
Soln ;
10
Equation in symbolic form is ,
(D2 + 1) y =cosecx
2
A E is D + 1 = 0
D=±1
C F = c1 cosx + c2 sinx
PI=
1X
f(D)
=
1 cosecx
D2 + 1
=
1 cosecx
(D +i)( D - i)
=
1
1
_ 1
2i ( D - i) (D +i )
= 1
1 cosecx
2i ( D - i)
Now
1 cosecx
D-i
cosecx
_ 1 cosecx
(D +i )
= eix ∫ cosecx e-ix dx
= eix ∫ cosecx ( cosx - isinx)dx
lll
rly
1 cosecx
D+ i
= eix ∫ (cotx – i) dx
= eix ( log sinx – ix )
= e-ix ( log sinx + ix )
P I = 1 [eix ( log sinx – ix ) _ e-ix ( log sinx + ix ) ]
2i
_
= log sinx
eix - e-ix
x
eix + e-ix
2i
2
= log sinx .sinx - x cosx
Hence the C S is y = c1 cosx + c2 sinx + log sinx .sinx - x cosx
6. Solve d2y - 2 dy +y = x ex sinx
dx2
dx
Soln:
Equation in symbolic form is ,
( D2 -2D +1)y = x ex sinx
2
A E is D -2D + 1 = 0
D = 1,1
(Partial fractions)
11
C F = ( c1 + c2 x ) ex
P I = 1 ex x sinx
( D -1)2
= ex 1 x sinx
( D +1-1 )2
= ex 1 x sinx
D2
= ex ∫ ∫ x sinx dx
= ex ∫
[ ∫ x sinx ] dx
= ex ∫
[
-x cosx + sinx] dx
= ex [-x sinx – cosx -cosx ]
= -ex [x sinx + 2cosx ]
Hence the C S is y = ( c1 + c2 x ) ex -ex [x sinx + 2cosx ]
7. Solve ( D – 2)2 y = 8 ( e2x + sin2x + x2 )
Soln :
A E is ( D – 2)2 = 0
D = 2,2
C F = ( c1 + c2 x ) e 2x
PI =
1
[8 ( e2x + sin2x + x2 ) ]
( D – 2)2
= 8
1 e2x + 1 sin2x + 1 x2
( D – 2)2 ( D – 2)2
( D – 2)2
Now 1 e 2x = x . 1 e 2x
( D – 2)2
2( D -2)
= x2 e 2x
12
2
1 sin2x =
1 sin2x
2
2–
( D – 2)
D 4D +4
= - 1 ∫ sin2x dx
4
= 1 cos2x
8
1 x2
( D – 2)2
=
1
x2
4(1 – D)2
2
= 1
4
= 1
4
1- D
2
-2
x2
1 + D + ¾ D2 + …….. x2
= 1 [x2 + 2x+ 3/2 ]
4
PI= 8
x2 e 2x + 1 cos2x + 1 [x2 + 2x+ 3/2 ]
2
8
8
= 4 x2 e 2x + cos2x + 2x2 + 4x+ 3
Hence the C S is y = ( c1 + c2 x ) e 2x + 4 x2 e 2x + cos2x + 2x2 + 4x+ 3
PROBLEMS
1.
Solve ( D 3 - 6D2 + 11D-6) y = e -2x + e-3x
2. Solve d2y - 4y = x sinhx
dx2
3. Solve d2y + a 2y = tanax
dx2
4. Solve ( D2 + 1) y = sinx sin2x
5. Solve ( D3 +3D2+2D)y = x2
6. Solve d2y + a 2y = secax
dx2
METHOD OF VARIATION OF PARAMETERS TO FIND THE P.I
13
Consider the second order linear differential equation with constant
coefficients
d2 y + a1 dy + a2 y = X……………………(1)
dx2
dx
Let its C.F be y = c1 y1 + c2 y2 so that y1 and y2 satisfy the equation
d2 y + a1 dy + a2 y = 0……………………(2)
dx2
dx
Now ,let us assume that the P.I of (1) is y = u y1 + v y2
where u and v are unknown functions of x.
u = - ∫y2 X dx and v =
W
W=
……………………..(3)
∫y1 X dx
W
y1 y2
y11 y 21
W is called the wronskian of y1 andy2
EXAMPLES
1. Apply the method of variation of parameters to solve d2 y + y = cosecx ?
dx2
Soln:
Given equation in symbolic form is (D2 + 1) y = cosecx
It’s A.E is D2 + 1 = 0 so that D= ±1i
C.F is y = c1 cosx + c2 sinx
Here y1 = cosx , y2 = sinx , X = cosecx
W=
=
y1 y2
y11 y 21
cosx
-sinx
sinx
cosx
= cos2x +sin2x = 1
P.I = - y1 ∫y2 X dx + y2 ∫y1 X dx
W
W
= - cosx ∫ sinx cosecx dx + sinx ∫ cosx cosecx dx
= - cosx ∫ sinx .1/sinx dx + sinx ∫ cosx .1/sinx dx
= - cosx ∫ dx + sinx ∫ cotx dx
14
= - cosx .x + sinx .log (sinx)
Hence the C.S is y = c1 cosx + c2 sinx - xcosx + sinx .log (sinx)
2. Apply the method of variation of parameters to solve d2 y -6dy + 9y = e3x ?
dx2 dx
x2
Soln:
Given equation in symbolic form is (D2 -6D + 9) y = e3x /x2
It’s A.E is D2 -6D + 9 = 0 so that D= 3,3
C.F is y = (c1+ c2 x) e3x = c1 e3x + c2 x e3x
Here y1 = e3x, y2 = xe3x , X = e3x /x2
W=
=
y1 y2
y11 y 21
e3x
3e3x
xe3x
3xe3x + e3x
= 3xe3x. e3x +e3x.e3x -3xe3xe3x
= e6x
P.I = - y1 ∫y2 X dx + y2 ∫y1 X dx
W
W
= - e3x ∫ (xe3x.e3x /x2 ) / e6xdx + xe3x ∫ (e3x.e3x /x2 ) / e6xdx
= - e3x ∫ 1/x dx + xe3x ∫ 1/x2 dx
= - e3x . logx + x e3x .-1/x
= -logx . e3x - e3x
= - e3x ( logx + 1)
Hence the C.S is y = (c1+ c2 x) e3x - e3x ( logx + 1).
PROBLEMS
1. Solve by the method of variation of parameters d2 y + 4y = 4 sec2 2x
dx2
2. Solve by the method of variation of parameters d2 y + y = cotx
dx2
3. Solve by the method of variation of parameters d2 y + y = secx
dx2
4. Solve by the method of variation of parameters ( D2 -3D +2) y = ex
1+ex
15
5. Solve by the method of variation of parameters d2 y - y = 2
dx2
1+ex
CAUCHY’S HOMOGENEOUS LINEAR EQUATION
An equation of the form
xn dny +a1 xn-1 dn-1y + a2 xn-2 dn-2y + …………………….. + an-1 x dy + an y = X
dxn
dxn-1
dxn-2
dx
……………………………………(1)
where ai’s are constants and X is a function of x, is called cauchy’s homogeneous
linear equation .
Such equations can be reduced to linear differential equations with constant coefficients
by the substitution x = ez or z = logx
so that dy = dy. dz
dx dz dx
= dy . 1
dz x
x. dy = dy = Dy where D = d
dx dz
dz
lllrly x2 d2 y = D (D-1)y
dx2
3 3
x d y = D (D-1)(D-2)y and so on .
dx3
Substituting these values in equation (1) we get a linear differential equation with
constant coefficients, which can be solved by the methods already discussed.
EXAMPLES
1.Solve ( x2 D2 + 3xD +5 )y = 10 – 4/x
Soln :
Given equation is a cauchy’s homogeneous linear equation .
Put x = ez , z = logx
so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y
Substituting these values in the given equation , it reduces to
D(D-1)y + 3Dy +5y = 10 – 4/ez
( D2 +2D+5) y = 10 – 4 e-z
which is a linear equation with constant coefficients.
Its A.E is D2 +2D+5 = 0
D = -1 ± 2i
C.F is y = e-z (c1 cos2z + c2 sin2z )
16
= 1/x (c1 cos(2logx) + c2 sin(2logx) )
P.I = 1/( D2 +2D+5) . (10 - 4 e-z )
= 1/( D2 +2D+5) .10 - 1/( D2 +2D+5) 4 e-z
= 10(1/( D2 +2D+5)) e0z – 4 (1/( D2 +2D+5)) e-z
= 2- e-z
= 2- (1/x )
Hence the C.S is y = 1/x (c1 cos(2logx) + c2 sin(2logx) ) + 2- (1/x ).
2. Solve x2 d2 y - x dy + y = logx
dx2
dx
Soln :
Given equation is a cauchy’s homogeneous linear equation .
Put x = ez , z = logx
so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y
Substituting these values in the given equation , it reduces to
D(D-1)y - Dy + y = z
( D2 -2D+1) y = z
which is a linear equation with constant coefficients.
Its A.E is D2 - 2D+1 = 0
D = 1,1
C.F is y = (c1 + c2z ) ez
= (c1 + c2 logx )x
P.I = 1/(D-1)2 .z
= (1-D)-2 z
= ( 1+ 2D + ………..) z
= ( 1+ 2D)z
=z+2
= logx + 2
Hence the C.S is y = (c1 + c2 logx )x + logx + 2.
3. Solve x2 d3 y - 4x d 2y +6dy = 4
dx3
dx2
dx
Soln :
Here the given equation is not a homogeneous linear equation but by
multiplying it by x it becomes
x3 d3 y/dx3 - 4x2 d 2y/dx2 +6xdy/dx = 4x……………..(1)
which is a cauchy’s homogeneous linear equation .
Put x = ez , z = logx
so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y, x3d3 y/dx3 = D (D-1)(D-2)y
Substituting these values in the equation (1), it reduces to
D(D-1)(D-2)y - 4D(D-1)y +6D y = 4ez
( D3 -7D2 +12 D) y = 4ez
which is a linear equation with constant coefficients.
17
Its A.E is D3 -7D2 +12 D = 0
D(D2 -7D +12) = 0
D = 0, 4,3
C.F is y = c1 e0z + c2e4z + c3 e3z
= c1+ c2x4 + c3 x3
P.I = ( 1/ D3 -7D2 +12 D ) 4ez
= 4ez /6
= 4x/6
=2x/3
Hence the C.S is y = c1+ c2x4 + c3 x3 + 2x/3.
PROBLEMS
1. Solve x3 d3 y +2x2 d 2y +2y = 10 (x +1/x).
dx3
dx2
2. Solve x2 d2 y - x dy - 3y = x2 logx.
dx2
dx
2 2
3. Solve x d y + x dy + y = logx. Sin(logx)
dx2
dx
2
4. Solve d y +1 dy = 2 + 5 logx.
dx2 x dx
5. Solve x2 d2 y - x dy + 4y = cos(logx) + x sin(logx).
dx2
dx
SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT
COEFFICIENTS
Differential equations in which there is one independent variable and two or
more than two dependent variables. Such equations are called simultaneous linear
equations. To solve such equations completely , we must have as many simultaneous
equations as the number of dependent variables. Here , we shall consider simultaneous
linear equations with constant coefficients only.
Let x, y be two dependent variables and t be the independent variable.
Consider the simultaneous equations
f1(D)x + f2 (D)y = T1 ……………(1) and ф1(D)x + ф2 (D)y = T2 ………………(2)
where D = d/dt and T1 , T2 are functions of t.
To eliminate y, operating on both sides of (1) by ф2 (D) and on both sides of (2) by
f2 (D) and subtracting , we get
( f1(D) ф2 (D)- ф1(D) f2 (D))x = ф2 (D)T1- f2 (D)T2 or f(D) x = T
Which is a linear equation in x and t and can be solved . substituting the value of x in
either (1) or (2) , we get the value of y.
EXAMPLES
18
1. Solve dx + 4x + 3y = t
dt
dy + 2x + 5y = et
dt
Soln :
Writing D for d the given equations become ( D +4) x + 3y = t …………..(1)
dt
2x + ( D+ 5)y = et ………….(2)
To eliminate y ,operating on both sides of (1) by ( D+5) and on both sides of (2) by 3 and
subtracting , we get
[( D +4) ( D+ 5) -6]x = (D +5)t – 3et
(D2 +9D+14)x = 1+5t -3et
Its AE is D2 +9D+14 = 0
D = -2, -7
CF = c1 e-2t + c2e-7t
1 (1+5t -3et )
D2 +9D+14
PI =
1 e0t + 1 5t
-3 et
2
D +9D+14 D +9D+14
D2 +9D+14
=
2
=
1
14
=
5t
14
+ 5 ( t – 9/14) - et
14
8
- 31 - et
196
8
x = c1 e-2t + c2e-7t + 5t
14
- 31 - et
196
8
Now dx/dt = -2c1 e-2t -7c2e-7t + 5 - et
14
8
Substituting the values of x and dx/dt in (1) we have
3y = t – dx/dt -4x
y = 1 [ -2c1 e-2t + 3c2e-7t - 3t
3
7
PROBLEMS
1.
Solve d2x + 4x + 5y = t 2
dt 2
+ 27 + 5et ]
98
8
19
d2y + 5x + 4y = t+1
dt 2
2. Solve tdx +y = 0
dt
t dy +x =0 given x(1) =1 , y(-1) = 0
dt
*********************************************************
20
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