(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Solve the following equation:

[tex](2x-y)dx+(4x+y-6)dy=0 [/tex]

2. Relevant equations

Solve for M and N as a linear system of equations; and

[tex]x_t = u + h [/tex]

[tex]y_t = v + k [/tex]

3. The attempt at a solution

M = 2x-y=0

N = 4x+y-6=0

2x=y

4x+2x=6

6x=6

x=1

y=2 ∴ [tex]x_t=u+1 \\ dx=du [/tex]

[tex]y_t=v+2 \\ dy=dv [/tex]

Substitute:

(2(u+1)-(v+2))du+(4(u+1)+(v+2)-6)dv=0

(2u-v)dw+(4u+v)dv=0 ;Homogeneous of degree one

u=vw du=vdw+wdv

(2vw-v)(vdw+wdv)+(4vw+v)dv=0

Multiply out:

[tex]\frac{(2w-1)dw}{(2w+1)(w+1)} + \frac{dv}{v} = 0 [/tex]

Then partial fractions for the first term:

[tex]\frac{(2w-1)}{(2w+1)(w+1)} = \frac{A}{2w+1} + \frac{B}{w+1}[/tex]

Solved:

A=3 B=-4

Substitute A and B:

[tex]\frac{3dw}{2w+1} + \frac{-4dw}{w+1}+ \frac{dv}{v} = 0 [/tex]

Integrated:

[tex]3ln|2w+1|-4ln|w+1|+ln|v|=C [/tex]

My question: How exactly does the book get this answer:[tex](x+y-3)^3=c(2x+y-4)^2[/tex]

Yes, I am bad at algebra.

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# Homework Help: Solving a Differential Equation with Coefficients Linear in 2 variables.

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