Question

Consider (4x + 3y)dx + (2xy)dy = 0
Find n, so that when the differential equation is multiplied by x^n, it becomes exact.

Answer 1

Math 201 UVic, eh?
If so you forgot the y^2, I'll solve assuming it's there.

Choose M = 4x + 3y^2
N = 2xy

Multiply both by x^n and let My = Nx
M = 4x^(n+1) + 3(x^n)y^2
N = 2(x^(n+1))y

My = 6(x^n)y = Ny = 2(n+1)(x^n)y
6(x^n)y = 2(n+1)(x^n)y

x^n and y cancels on both sides
6 = 2(n+1)
3 = n + 1
n = 2

Just multiply the DE by x^2 and solve like any other exact differential equation.

Answer 2

You should get (x^3)(y^2) + x^4 = c
for an answer.

Answer 3

Have you figured out 5 and 6 yet?

Answer 4

haha yeah 201. yeah its not bad once you know the typo. working on 5 and 6 now, have no idea for 5