# General Solution Differential Equation (integrating factors)

## Homework Statement

find the general solution using integrating factors.
xy' - 2y = 3x

## The Attempt at a Solution

x(dy/dx) = 3x + 2y
x*dy = (3x + 2y)dx
(-3x + 2y)dx + xdy = 0

My = -3x + 2
Nx = 1
Not Exact (hence the use of integrating factors)

μ(x)(-3x + 2y)dx + μ(x)xdy = 0

Differentiating with respect to y for M and x for N
μ(x)(-3x + 2) = μ'(x)x + μ(x)

trying to simplify
-3xμ(x) + 2μ(x) - μ(x) = μ'(x)
-3xμ(x) - μ(x) = μ'(x)

-μ(x)(3x - 1) = μ'(x)

I know I need to solve for μ(x) to multiply M and N by, I am not sure if I have done everything correct up until now. I assume that I possibly just need a little help with the algebra to be able to set up an integral to solve for μ(x) but I am not sure.

Any help would be greatly appreciated, I'm not sure if I missed something from the beginning or am almost there. Thanks ahead of time!

Last edited:

Just realized I did My wrong, re-trying now.

Second attempt:

Equations:
F = e∫P(x)dx
P(x) = 1/N*(My - Nx)
ψx = M ψy = N

Attempt:
xy' - 2y = 3x
-2y - 3x + xy' = 0
My = -2 Nx = 1
1/N(My - Nx)
1/x(-2 - 1)
3x-1
∫3x-1dx → -3ln|x|
e-3ln|x| = -3x

integrating factor = -3x

(-3x)(-2y - 3x) + (-3x)xy' = 0
(-6xy - 9x2) - (3x2)y' = 0
My = -6x Ny = -6x EXACT

Integrating M to get ψ(x,y)
∫ψxdx = ∫-6xy - 9x2dx
ψ(x,y) = -3x2y - 3x3 + f(y)

Differentiate with respect to y to get N
ψy = -3x2 + f'(y)
where N = ψy
-3x2 = -3x2 + f'(y)
0 = f'(y) → ∫f'(y)dy = ∫dx
f(y) = c

ψ = -3x2y - 3x3 + c

I think I did everything right. I'm not sure if this is the correct answer or not any help or checking of my work would be appreciated thanks.