Solve by using I.F method?

Question

(2xy^4e^y + 2xy^3 + y)dx + (x^2y^4e^y - x^2y^2-3x)dy = 0

Answer 1

(2xy⁴ e^y + 2xy³ + y) dx + (x²y⁴ e^y - x²y² - 3x) dy = 0

First, check to see whether this equation is exact. Let P = (2xy⁴ e^y + 2xy³ + y) and Q = (x²y⁴ e^y - x²y² - 3x):

∂P/∂y = 8xy³ e^y + 2xy⁴ e^y + 6xy² + 1
∂Q/∂x = 2x⁴ e^y - 2xy² - 3

Not even close to being exact. So we need an integrating factor, u(x, y), such that if we multiply both sides by u, the equation will be exact. Meaning:

∂(uP)/∂y = ∂(uQ)/∂x, and thus:

∂u/∂y P + u ∂P/∂y = ∂u/∂x Q + u ∂Q/∂x

Now, this PDE will be a PITA to solve if u is a function of both variables. But suppose that u is a function of y only: then we have ∂u/∂x = 0, so solving for (∂u/∂y)/u:

∂u/∂y P = u (∂Q/∂x - ∂P/∂y)
(∂u/∂y)/u = (∂Q/∂x - ∂P/∂y)/P

Obviously, this means that (∂Q/∂x - ∂P/∂y)/P will be a function of y only. Conversely, if u is a function of x only, a similar procedure reveals that (∂P/∂y - ∂Q/∂x)/Q will be a function of x only. In either case, it will be helpful to know what ∂Q/∂x - ∂P/∂y is, so computing it:

∂Q/∂x - ∂P/∂y
(2x⁴ e^y - 2xy² - 3) - (8xy³ e^y + 2xy⁴ e^y + 6xy² + 1)
- 8xy³ e^y - 8xy² - 4
- 4/y (2xy⁴ e^y + 2xy³ + y)
-4/y P

So indeed, (∂Q/∂x - ∂P/∂y)/P = -4/y and so is a function of y only. So u is a function of y only, and we have:

(∂u/∂y)/u = (∂Q/∂x - ∂P/∂y)/P = -4/y
ln |u| = ∫-4/y dy = -4 ln |y|
u = y⁻⁴.

So multiplying by the integrating factor:

(2x e^y + 2xy⁻¹ + y⁻³) dx + (x² e^y - x²y⁻² - 3xy⁻⁴) dy = 0

And if we redefine P = (2x e^y + 2xy⁻¹ + y⁻³) and Q = (x² e^y - x²y⁻² - 3xy⁻⁴), and check their partial derivatives:

∂P/∂y = 2x e^y - 2xy⁻² - 3y⁻⁴
∂Q/∂x = 2x e^y - 2xy⁻² - 3y⁻⁴

So this equation is now exact, and we can get on with solving it. We are looking for a function ψ such that ∂ψ/∂x = P and ∂ψ/∂y = Q. So solving these equations:

∂ψ/∂x = 2x e^y + 2xy⁻¹ + y⁻³
ψ = ∫2x e^y + 2xy⁻¹ + y⁻³ dx
ψ = x²e^y + x²y⁻¹ + xy⁻³ + C(y)
∂ψ/∂y = x²e^y - x²y⁻² - 3xy⁻⁴ + ∂C/∂y
x² e^y - x²y⁻² - 3xy⁻⁴ = x²e^y - x²y⁻² - 3xy⁻⁴ + ∂C/∂y
∂C/∂y = 0
C(y) = K (K is constant)
ψ = x²e^y + x²y⁻¹ + xy⁻³ + K

Thus we have that:

x²e^y + x²y⁻¹ + xy⁻³ = K

Finding an explicit solution is not practical, so we'll leave this as an implicit solution. So we are done.