Solving a Differential Equation with Coefficients Linear in 2 variables.

  • Thread starter Kiziaru
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  • #1
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Homework Statement


Solve the following equation:

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

Homework 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]

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.
 
Last edited:

Answers and Replies

  • #2
DryRun
Gold Member
838
4
First, you must re-arrange your differential equation in this form:
[tex]\frac{dy}{dx}=\frac{y-2x}{y+4x−6}=\frac{\frac{y}{x}-2}{\frac{y}{x}+4−\frac{6}{x}}[/tex]
Since the equation above is not a function of [itex]\frac{y}{x}[/itex] only, it is an inhomogeneous equation. Hence, we seek a translation of axes: x=X+h and y=Y+k
[tex]\frac{dY}{dX}=\frac{Y-2X+(k-2h)}{Y+4X+(k+4h-6)}[/tex]
[tex]k-2h=0
\\k+4h-6=0[/tex]
Solving the above system for h and k give: h=1 and k=2
So, we let x=X+1 and y=Y+2
[tex]\frac{dY}{dX}=\frac{Y-2X}{Y+4X}=\frac{\frac{Y}{X}-2}{\frac{Y}{X}+4}[/tex]
Since the above equation is now homogeneous, let [itex]\frac{Y}{X}=v[/itex]
Then, [tex]\frac{dY}{dX}=v+X\frac{dv}{dX}[/tex]
Solving, we get:
[tex]\ln X= -2v + \ln (v+2)^2 +\ln (v+1)^3[/tex]
[tex]2v=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
[tex]2\frac{Y}{X}=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
[tex]Y=\frac{X}{2} \ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
Finally, replace v, Y and X with the original y and x.
 
Last edited:
  • #3
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Then, [tex]\frac{dY}{dX}=v+X\frac{dv}{dX}[/tex]
Solving, we get:
[tex]\ln X= -2v + \ln (v+2)^2 +\ln (v+1)^3[/tex]
[tex]2v=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
[tex]2\frac{Y}{X}=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
[tex]Y=\frac{X}{2} \ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
Finally, replace v, Y and X with the original y and x.

It's here that I am getting lost.

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

Which does not integrate into something my textbook has.
 
  • #4
6
0
Never mind, I figured it out. It only took me a day. :P

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

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