# Homogeneous DE with Transformation

• Jan 19th 2011, 10:32 AM
bugatti79
Homogeneous DE with Transformation
All,

I don know how to solve this linear ordinary DE given the transfomation v=x+y. I havent come across this before. I have solved other ones involving y=vx where v is a function of x, thereby usnig the product rule to get dy/dx etc

$\displaystyle \frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}$

Any clues will be appreciated.
Thanks
• Jan 19th 2011, 11:13 AM
TheEmptySet
Quote:

Originally Posted by bugatti79
All,

I don know how to solve this linear ordinary DE given the transfomation v=x+y. I havent come across this before. I have solved other ones involving y=vx where v is a function of x, thereby usnig the product rule to get dy/dx etc

$\displaystyle \frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}$

Any clues will be appreciated.
Thanks

You are almost there but you need to find $\displaystyle \frac{dy}{dx}$ is terms of $\displaystyle v$.

$\displaystyle \displaystyle v=x+y \implies \frac{dv}{dx}=1+\frac{dy}{dx} \implies \frac{dy}{dx}=\frac{dv}{dx}-1$

Putting this into what you have above gives

$\displaystyle \displaystyle \frac{dv}{dx}-1=-\frac{v}{3v-4}$

can you finish from here?
• Jan 19th 2011, 01:32 PM
bugatti79
Quote:

Originally Posted by TheEmptySet
You are almost there but you need to find $\displaystyle \frac{dy}{dx}$ is terms of $\displaystyle v$.

$\displaystyle \displaystyle v=x+y \implies \frac{dv}{dx}=1+\frac{dy}{dx} \implies \frac{dy}{dx}=\frac{dv}{dx}-1$

Putting this into what you have above gives

$\displaystyle \displaystyle \frac{dv}{dx}-1=-\frac{v}{3v-4}$

can you finish from here?

ok, from your last line,I rearrange, separate variables and integrate

$\displaystyle \int \frac{3v-4}{2v-4}dv=\int dx$ from this I get

$\displaystyle \int \frac{3}{2} dv+\int \frac{1}{v-2}dv=\int dx$ therefore, after some manipulation, giving

$\displaystyle x+3y+2ln((x+y)-2)=-2c$

The answer in the book for the LHS is the same and for the RHS its A, so I could call -2c =A, right?...
• Jan 19th 2011, 01:34 PM
TheEmptySet
Quote:

Originally Posted by bugatti79
ok, from your last line,I rearrange, separate variables and integrate

$\displaystyle \int \frac{3v-4}{2v-4}dv=\int dx$ from this I get

$\displaystyle \int \frac{3}{2} dv+\int \frac{1}{v-2}dv=\int dx$ therefore, after some manipulation, giving

$\displaystyle x+3y+2ln((x+y)-2)=-2c$

The answer in the book for the LHS is the same and for the RHS its A, so I could call -2c =A, right?...

Yes you can relable your constants.
• Jan 19th 2011, 01:37 PM
bugatti79
can I retract the double reply...I sent it by accident? I dont want to violate the forum rules?