• 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