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Hey guys, I've been stuck on this problem for a good hour... I have no idea how to finish it up.

Solve: dy/dx = 4y-3x / (2x-y)

Use homogeneous equations method.

Answer : |y - x| = c|y + 3x|

dy/dx = 4y-3x / (2x-y)

dy/dx = 4y-3x / (2x-y) * [(1/x) / (1/x)]

dy/dx = (4y/x)-3 / (2-(y/x))

let v = y/x

y = v(x) x

dy/dx = x dv/dx + v

x dv/dx + v = 4v-3 / (2-v)

x dv/dx = [4v-3 / (2-v)] - v

x dv/dx = 4v-3 -[2v-v

x dv/dx = 2v-3+v

dx/x = [(-v + 2) / (v^2 + 2v -3)] dv

dx/x = [(-5/4)(1/(v+3)) + (1/4)(1/(v-1))] dv (partial fractions)

Integrating both sides:

ln |x| = (1/4) ln |v-1| - (5/4) ln |v+3| + c

ln |x| = (1/4) [ln |v-1| - 5 ln |v+3| + c]

|x| = (1/4) exp {[ln |v-1| - 5 ln |v+3| + c]}

|x| = c |v-1| |v+3|

c|v+3|

|(y/x)-1| / |x| = c|(y/x)+3|

Much appreciated.

## Homework Statement

Solve: dy/dx = 4y-3x / (2x-y)

Use homogeneous equations method.

## Homework Equations

Answer : |y - x| = c|y + 3x|

^{5}(also y = -3x)## The Attempt at a Solution

dy/dx = 4y-3x / (2x-y)

dy/dx = 4y-3x / (2x-y) * [(1/x) / (1/x)]

dy/dx = (4y/x)-3 / (2-(y/x))

let v = y/x

y = v(x) x

dy/dx = x dv/dx + v

x dv/dx + v = 4v-3 / (2-v)

x dv/dx = [4v-3 / (2-v)] - v

x dv/dx = 4v-3 -[2v-v

^{2}] / (2-v)x dv/dx = 2v-3+v

^{2}] / (2-v)dx/x = [(-v + 2) / (v^2 + 2v -3)] dv

dx/x = [(-5/4)(1/(v+3)) + (1/4)(1/(v-1))] dv (partial fractions)

Integrating both sides:

ln |x| = (1/4) ln |v-1| - (5/4) ln |v+3| + c

ln |x| = (1/4) [ln |v-1| - 5 ln |v+3| + c]

|x| = (1/4) exp {[ln |v-1| - 5 ln |v+3| + c]}

|x| = c |v-1| |v+3|

^{-5}c|v+3|

^{5}= |v-1|/|x||(y/x)-1| / |x| = c|(y/x)+3|

^{5}*Answer from book : |y - x| = c|y + 3x|*^{5}(also y = -3x)Much appreciated.

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