2023 APM1514 Assignment 2 Memo
Question 1
1.1
dy 2 y−t 1
= (t + 1)e
dt cos(y)
2 −t y 1
= (t + 1)e e
cos(y)
cos(y) · e−y dy = (t2 + 1)e−t dt
Thus, the equation is separable.
1.2
dy
2x2 = x2 + y 2
dx
dy x2 + y 2
=
dx 2x2
This is not separable. It is a homogeneous equation and can be solved with
substitution. Let y = vx, hence v = xy .
Using the product rule:
dy dx dv
=v· +x·
dx dx dx
Substituting into the equation:
dv 1 + v2
x· +v =
dx 2
dv (v − 1)2
x· =
dx 2
Separate variables:
2 1
2
dv = dx
(v − 1) x
Integrate: Z Z
2 1
dv = dx
(v − 1)2 x
1
Question 1
1.1
dy 2 y−t 1
= (t + 1)e
dt cos(y)
2 −t y 1
= (t + 1)e e
cos(y)
cos(y) · e−y dy = (t2 + 1)e−t dt
Thus, the equation is separable.
1.2
dy
2x2 = x2 + y 2
dx
dy x2 + y 2
=
dx 2x2
This is not separable. It is a homogeneous equation and can be solved with
substitution. Let y = vx, hence v = xy .
Using the product rule:
dy dx dv
=v· +x·
dx dx dx
Substituting into the equation:
dv 1 + v2
x· +v =
dx 2
dv (v − 1)2
x· =
dx 2
Separate variables:
2 1
2
dv = dx
(v − 1) x
Integrate: Z Z
2 1
dv = dx
(v − 1)2 x
1
