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Homogeneous DE Problem

  • Thread starter snowJT
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118
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1. Homework Statement

I'm told that this is homogenous
[tex](x^2-xy)y'+y^2 = 0[/tex]

2. The attempt at a solution

This is going to be very painful for me to type out...

[tex](x^2-xy)\frac{dy}{dx} = -y^2[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

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

[tex]v+x\frac{dv}{dx} = \frac{-x^2v^2}{x^2(1-v)}[/tex]

[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]

[tex]\frac{v+1-v}{-v^2}dv = \frac {dx}{x}[/tex]

[tex]\frac{dv}{-v^2} = \frac {dx}{x}[/tex]

[tex]\int-v^-^2dv = \int\frac {dx}{x}[/tex]

[tex]v^-^1 =ln|x|+C[/tex]

[tex]\frac{y^-^1}{x^-^1} = ln|x|+C[/tex]

[tex]y^-^1 = x ln|x|+C[/tex]

[tex]\frac{1}{y} = x ln|x|+C[/tex]

[tex]\frac{1}{y} = \frac {ln|x|+C}{x}[/tex]

[tex] 0 = \frac {y ln|x|+C}{x}[/tex]

[tex] -Cx = y ln|x|[/tex]

But the answer is...

[tex] xln|y| - y = Cx[/tex]
 
Last edited:

Answers and Replies

cristo
Staff Emeritus
Science Advisor
8,056
72
1. Homework Statement

I'm told that this is homogenous
[tex](x^2-xy)y'+y^2 = 0[/tex]

2. The attempt at a solution

This is going to be very painful for me to type out...

[tex](x^2-xy)\frac{dy}{dx} = -y^2[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

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

[tex]v+x\frac{dv}{dx} = \frac{-x^2v^2}{x^2(1-v)}[/tex]

[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]
Try going from here again. Add another line in if you need to. Put all the v's over a common denominator before you multiply by dx.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
897
1. Homework Statement

I'm told that this is homogenous
[tex](x^2-xy)y'+y^2 = 0[/tex]

2. The attempt at a solution

This is going to be very painful for me to type out...

[tex](x^2-xy)\frac{dy}{dx} = -y^2[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

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

[tex]v+x\frac{dv}{dx} = \frac{-x^2v^2}{x^2(1-v)}[/tex]

[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]
[tex]\frac{-v^2}{1-v}-v= \frac{-v^2-v+v^2}{1-v}= \frac{-v}{1-v}[/tex]
The left hand side in your next line is wrong.

[tex]\frac{v+1-v}{-v^2}dv = \frac {dx}{x}[/tex]

[tex]\frac{dv}{-v^2} = \frac {dx}{x}[/tex]

[tex]\int-v^-^2dv = \int\frac {dx}{x}[/tex]

[tex]v^-^1 =ln|x|+C[/tex]

[tex]\frac{y^-^1}{x^-^1} = ln|x|+C[/tex]

[tex]y^-^1 = x ln|x|+C[/tex]

[tex]\frac{1}{y} = x ln|x|+C[/tex]

[tex]\frac{1}{y} = \frac {ln|x|+C}{x}[/tex]

[tex] 0 = \frac {y ln|x|+C}{x}[/tex]

[tex] -Cx = y ln|x|[/tex]

But the answer is...

[tex] xln|y| - y = Cx[/tex]
 
118
0
Oh I understand the error, thanks, I'll work it out from there
 
118
0
I've messed up again...

[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]
[tex]\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}[/tex]
[tex]\frac{v-v^2}{-v^2}dv = \frac{dx}{(x)}[/tex]
[tex]\int v^-^1-1 dv = \int \frac{dx}{(x)}[/tex]
[tex]ln|v|-x = ln|x|+C[/tex]
[tex]ln|\frac {y}{x}|-x = ln|x|+C[/tex]

and no need to go any farter.. I know its wrong...
 
118
0
is it cus this line

[tex]\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}[/tex]

should look like this?

[tex]\frac{v(1-v)}{-v^2} + (1-v)dv = \frac{dx}{(x)}[/tex]
 
cristo
Staff Emeritus
Science Advisor
8,056
72
Try going from this line. Simplify before you multiply by dx.

[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]

[tex]x\frac{dv}{dx}=\frac{-v^2}{(1-v)}-v=\frac{-v^2-v(1-v)}{(1-v)} [/tex]

Simplify from here, and then carry on!
 

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