Math 2280
Quiz 2
1. Find the general solution of the equation
dy
= y sin x.
dx
It is clear that this equation is seperable, so we rewrite it as
dy
= sin x dx
y
and integrate to obtain
Z
dy
=
y
Z
sin x dx ⇒ ln y = − cos x + C ⇒ y = De− cos x
for any choice of constant D.
2. Find the solution of the differential equation
dy
= 2xy 2 + 3x2 y 2
dx
satisfying the initial condition y(1) = −1.
Again we have a separable equation, which is more easily seen when we write it as
dy
= (2x + 3x2 )y 2 .
dx
Once we rearrange and integrate we have
Z
Z
−1
dy
= (2x + 3x2 ) dx ⇒ −y −1 = x2 + x3 + C ⇒ y = 2
.
2
y
x + x3 + C
The initial condition implies that
−1 = y(1) =
−1
⇒ 2 + C = 1 ⇒ C = −1,
2+C
so our particular solution is
y(x) =
−1
.
x2 + x3 − 1
1