Challenging First Order D.E.
Paul Skoufranis
Due Thursday, February 12th
Consider the differential equation
(4xy 4 + 3y) + (2x2 y 3 − x)
dy
= 0.
dx
You may check for yourself that this differential equation is not separable, linear, or exact. We will try to
make this differential equation exact using an integrating factor I(x, y).
a) By multiplying both sides of the differential equation by I(x, y), the differential equation becomes
(4xy 4 + 3y)I(x, y) + (2x2 y 3 − x)I(x, y)
dy
= 0.
dx
Using the cross-partial condition, show that this differential equation is exact provided
(2x2 y 2 − x)
∂I
∂I
− (4xy 4 + 3y)
= (12xy 3 + 4)I.
∂x
∂y
Unfortunately, the above differential equation is a partial differential equation which we do not know
how to solve. Also, trying to make I just a function of x or just a function of y does not help as setting the
appropriate partial derivatives to zero does not eliminate one variable from the differential equation.
Instead, we will make I a function of u = xy ; that is, I will be a function of the ratio of x and y.
b) Using the facts that
∂u ∂I
∂I
=
∂y
∂y ∂u
and
∂I
∂u ∂I
=
,
∂x
∂x ∂u
show that the above partial differential equation becomes
x ∂I
2 2
6x y + 2
= (12xy 3 + 4)I.
y ∂u
c) Using the fact that
x
3x y +
y
2 2
=
x
3xy 3 + 1
y
reduce the above differential equation to one involving only I and u. Solve for I first in terms of u, and then
in terms of x and y.
d) Solve the original differential equation using the integrating factor from part (c).