Question

Test for exactness. If exact, solve it directly. Otherwise, use integrating factors to solve it. Solve the IVP (if given). 2xy + (x^2) y' = 0
sin(x) cos(y) + cos(x) sin(y) y' = 0
(x^2) + (y^2) - 2xyy' = 0
e^(2x).(2 cos(y) - sin(y) y') = 0, where y(0) = 0​

Answer 1

• 2xy + x ² y' = 0

This DE is exact, since

∂(2xy)/∂y = 2x

∂(x ²)/∂x = 2x

are the same. Then there is a solution of the form f(x, y) = C such that

∂f/∂x = 2xy   ==>   f(x, y) = x ² y + g(y)

∂f/∂y = x ² = x ² + dg/dy   ==>   dg/dy = 0   ==>   g(y) = C

==>   f(x, y) = x ² y = C

• sin(x) cos(y) + cos(x) sin(y) y' = 0

is also exact because

∂(sin(x) cos(y))/∂y = -sin(x) sin(y)

∂(cos(x) sin(y))/∂x = -sin(x) sin(y)

Then

∂f/∂x = sin(x) cos(y)   ==>   f(x, y) = -cos(x) cos(y) + g(y)

∂f/∂y = cos(x) sin(y) = cos(x) sin(y) + dg/dy   ==>   dg/dy = 0   ==>   g(y) = C

==>   f(x, y) = -cos(x) cos(y) = C

• x ² + y ² - 2xyy' = 0

is not exact:

∂(x ² + y ²)/∂x = 2x

∂(-2xy)/∂y = -2x

So we look for an integrating factor µ(x, y) such that

µ (x ² + y ²) - 2µxyy' = 0

becomes exact, which would require that these be equal:

∂(µ (x ² + y ²))/∂y = (x ² + y ²) ∂µ/∂y + 2µy

∂(-2µxy)/∂x = -2xy ∂µ/∂x - 2µy

Observe that if µ(x, y) = µ(x), then ∂µ/∂y = 0 and ∂µ/∂x = dµ/dx, so we would have

2µy = -2xy dµ/dx - 2µy

==>   -2xy dµ/dx = 4µy

==>   dµ/µ = -2/x dx

Integrating both sides gives

∫ dµ/µ = ∫ -2/x dx   ==>   ln|µ| = -2 ln|x|   ==>   µ = 1/x ²

So in the modified DE, we have

(1 + y ²/x ²) - 2y/x y' = 0

which is now exact and ready to solve, since

∂(1 + y ²/x ²)/∂y = 2y/x ²

∂(-2y/x)/∂x = 2y/x ²

We get

∂f/∂x = 1 + y ²/x ²   ==>   f(x, y) = x - y ²/x + g(y)

∂f/∂y = -2y/x = -2y/x + dg/dy   ==>   dg/dy = 0   ==>   g(y) = C

==>   f(x, y) = x - y ²/x = C

• exp(2x) (2 cos(y) - sin(y) y' ) = 0

is exact, since

∂(2 exp(2x) cos(y))/∂y = -2 exp(2x) sin(y)

∂(-exp(2x) sin(y))/∂x = -2 exp(2x) sin(y)

Then

∂f/∂x = 2 exp(2x) cos(y)   ==>   f(x, y) = exp(2x) cos(y) + g(y)

∂f/∂y = -exp(2x) sin(y) = -exp(2x) sin(y) + dg/dy   ==>   dg/dy = 0   ==>   g(y) = C

==>   f(x, y) = exp(2x) cos(y) = C

Given that y = 0 when x = 0, we find that

C = exp(0) cos(0) = 1

so that the particular solution is

exp(2x) cos(y) = 1