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: $$(x+y-3)^3=c(2x+y-4)^2$$Yes, 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
$$\frac{dY}{dX}=\frac{Y-2X+(k-2h)}{Y+4X+(k+4h-6)}$$
$$k-2h=0 \\k+4h-6=0$$
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.