Download Problem set 2

MATH 225 / MTHE 225
Problem Set 2
1. Solve the differential equation: yy 0 = x(1 + y 2 ).
2. Solve the differential equation: y 0 =
(1 − y)6
.
5
3. Solve the differential equation: y 0 = 6(1 + x)5 .
4. Solve the differential equation: y 0 = y(y − 1).
5. Solve the differential equation: y 0 = 2xex
2 −y
.
6. Solve the initial value problem (IVP): y 0 = ln (x)(1 + y 2 );
7. Solve the initial value problem (IVP): y 0 = 8e4x (1 + y 2 );
8. Solve the differential equation: y 0 = 1 + x + y + xy.
1
y(1) = 5.
y(0) = 2.
Solution
1.
(Separable DE)
y
dy
= x(1 + y 2 ) ⇒
yy = x(1 + y ) ⇒ y
dy = x dx ⇒
dx
1 + y2
0
2
Z
y
dy =
1 + y2
Z
x dx
For the integral on the left, we need to use u-substitution with u = 1 + y 2 .
1
That means, du = 2y dy ⇒ du = y dy.
2
Z
Z
y
du
1
1
1
So,
=
ln
u
=
ln (1 + y 2 ).
dy
=
2
1+y
2
u
2
2
Now going back to our ODE we get:
x2
1
ln (1 + y 2 ) =
+ C (implicit solution)
2
2
2
⇒ ln (1 + y 2 ) = x2 + 2C ⇒ 1 + y 2 = ex +2C ⇒ y 2 = ex
p
⇒ y = ± ex2 +2C − 1 (explicit solution).
2 +2C
−1
2.
(Separable DE)
Z
Z
dy
(1 − y)6
dy
1
dy
1
(1 − y)6
0
⇒
=
⇒
= dx ⇒
=
dx
y =
5
dx
5
(1 − y)6
5
(1 − y)6
5
Z
Z
1
−6
⇒ (1 − y) dy =
dx
5
(1 − y)−5
1
⇒−
= x + C (implicit solution)
−5
5
1
1
1
5
⇒
=
x
+
5C
⇒
(1
−
y)
=
⇒
1
−
y
=
(1 − y)5
x + 5C
(x + 5C)1/5
1
⇒y =1−
(explicit solution).
(x + 5C)1/5
3.
(Separable DE)
Z
Z
dy
5
5
y = 6(1 + x) ⇒
= 6(1 + x) ⇒ dy = 6(1 + x) dx ⇒
dy = 6 (1 + x)5 dx
dx
6
(1 + x)
+ C ⇒ y = (1 + x)6 + C (explicit solution).
⇒y=6
6
0
4.
5
(Separable DE)
Z
Z
1
dy
1
0
y = y(y − 1)
= y(y − 1) ⇒
dy = dx ⇒
dy =
dx
dx
y(y − 1)
y(y − 1)
(Now “partial fraction decomposition” can be used to calculate the integral of the left hand
side, but we can avoid it here.)
2
Z
Z
Z
1
y − (y − 1)
⇒
dy =
dx ⇒
dy =
dx
y(y − 1)
y(y − 1)
Z Z
Z Z
y
(y − 1)
1
1
−
dy =
dx ⇒
−
dy =
dx
⇒
y(y − 1) y(y − 1)
y−1 y
Z
⇒ ln (y − 1) − ln y = x + C (implicit solution)
y−1
1
1
y−1
=x+C ⇒
= ex+C ⇒ 1 − = ex+C ⇒ = 1 − ex+C
⇒ ln
y
y
y
y
1
⇒y=
(explicit solution).
1 − ex+C
5.
(Separable DE)
2
dy
2xex
dy
2
2
= 2xex −y ⇒
=
y = 2xe
⇒
⇒ ey dy = 2xex dx
y
dx
e
Z
Z dx
y
x2
⇒ e dy = 2xe dx
0
x2 −y
2
For the integral on the right, we need to use u-substitution with u = ex .
2
That means, du = 2xex dx
Z
Z
2
x2
So, 2xe =
du = u = ex .
Now going back to our ODE we get:
Z
Z
2
2
y
e dy = 2xex dx ⇒ ey = ex + C (implicit solution).
2
⇒ y = ln (ex + C) (explicit solution).
6.
(Separable DE)
dy
dy
y = ln (x)(1+y ) ⇒
= ln (x)(1+y 2 ) ⇒
= ln x dx ⇒
dx
(1 + y 2 )
0
2
Z
dy
=
(1 + y 2 )
Z
ln x dx
For the integral on the right, we need to use integration by parts with u = ln x and dv = dx.
1
That means du = dx and v = x.
x
Z
Z
Z
Z
1
So ln x dx = uv − v du = (ln x)x − x dx = x(ln x) −
dx = x(ln x) − x.
x
Now going back to our ODE we get:
Z
Z
dy
= ln x dx ⇒ tan−1 y = x(ln x) − x + C (implicit solution).
(1 + y 2 )
⇒ y = tan (x(ln x) − x + C) (explicit solution).
3
But we need y(1) = 5.
That means, 5 = tan (1 ln (1) − 1 + C) ⇒ 5 = tan (C − 1) ⇒ C = 1 + tan−1 5. Plugging
this value of C in our explicit solution we get the solution of the IVP as:
y = tan x(ln x) − x + 1 + tan−1 5 .
7.
(Separable DE)
Z
Z
dy
dy
dy
4x
2
4x
y = 8e (1 + y ) ⇒
= 8e (1 + y ) ⇒
= 8e dx ⇒
= 8e4x dx
2
2
dx
1+y
1+y
4x
e
+ C ⇒ tan−1 y = 2e4x + C (implicit solution).
⇒ tan−1 y = 8
4
⇒ y = tan 2e4x + C (explicit solution).
0
4x
2
But we need y(0) = 2.
That means, 2 = tan (2e0 + C) ⇒ 2 = tan (2 + C) ⇒ C = tan−1 2 − 2. Plugging this value
of C in our explicit solution we get the solution of the IVP as:
y = tan 2e4x + tan−1 2 − 2 .
8.
(Separable DE)
dy
y 0 = 1 + x + y + xy = (1 + x) + (1 + x)y = (1 + x)(1 + y) ⇒
= (1 + x)(1 + y)
dx
Z
Z
dy
dy
= (1 + x) dx ⇒
= (1 + x) dx
⇒
1+y
1+y
x2
⇒ ln (1 + y) = x +
+ C (implicit solution)
2
x2
x2
⇒ 1 + y = ex+ 2 +C ⇒ y = ex+ 2 +C − 1 (explicit solution)
4