Solving a Differential Equation with Coefficients Linear in 2 variables.

[Kiziaru]

Homework Statement

Solve the following equation:

(2x-y)dx+(4x+y-6)dy=0

Homework Equations

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

x_t = u + h
y_t = v + k

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 ∴ x_t=u+1 \\ dx=du
y_t=v+2 \\ dy=dv

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:

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

Then partial fractions for the first term:

\frac{(2w-1)}{(2w+1)(w+1)} = \frac{A}{2w+1} + \frac{B}{w+1}

Solved:

A=3 B=-4

Substitute A and B:

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

Integrated:

3ln|2w+1|-4ln|w+1|+ln|v|=C

My question: How exactly does the book get this answer: (x+y-3)^3=c(2x+y-4)^2Yes, I am bad at algebra.

 

[DryRun]

First, you must re-arrange your differential equation in this form:
\frac{dy}{dx}=\frac{y-2x}{y+4x−6}=\frac{\frac{y}{x}-2}{\frac{y}{x}+4−\frac{6}{x}}
Since the equation above is not a function of \frac{y}{x} only, it is an inhomogeneous equation. Hence, we seek a translation of axes: x=X+h and y=Y+k
\frac{dY}{dX}=\frac{Y-2X+(k-2h)}{Y+4X+(k+4h-6)}
k-2h=0<br /> \\k+4h-6=0
Solving the above system for h and k give: h=1 and k=2
So, we let x=X+1 and y=Y+2
\frac{dY}{dX}=\frac{Y-2X}{Y+4X}=\frac{\frac{Y}{X}-2}{\frac{Y}{X}+4}
Since the above equation is now homogeneous, let \frac{Y}{X}=v
Then, \frac{dY}{dX}=v+X\frac{dv}{dX}
Solving, we get:
\ln X= -2v + \ln (v+2)^2 +\ln (v+1)^3
2v=\ln \frac{(v+2)^2 (v+1)^3}{X}
2\frac{Y}{X}=\ln \frac{(v+2)^2 (v+1)^3}{X}
Y=\frac{X}{2} \ln \frac{(v+2)^2 (v+1)^3}{X}
Finally, replace v, Y and X with the original y and x.

 

[Kiziaru]

Then, \frac{dY}{dX}=v+X\frac{dv}{dX}
Solving, we get:
\ln X= -2v + \ln (v+2)^2 +\ln (v+1)^3
2v=\ln \frac{(v+2)^2 (v+1)^3}{X}
2\frac{Y}{X}=\ln \frac{(v+2)^2 (v+1)^3}{X}
Y=\frac{X}{2} \ln \frac{(v+2)^2 (v+1)^3}{X}
Finally, replace v, Y and X with the original y and x.

It's here that I am getting lost.

I have also have \frac{dY}{dX}=v+X\frac{dv}{dX} and set it equal to \frac{v-2}{v+4} but when I separate the variables I get \frac{(v+4)dv}{v^2+3v+2}=\frac{-dX}{X}

Which does not integrate into something my textbook has.

 

[Kiziaru]

Never mind, I figured it out. It only took me a day. :P

Also, thanks for everything. You were a huge help.