Homogeneous DE with Transformation

**bugatti79** · Jan 19th 2011, 11:32 AM

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

$<br /> \frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}<br />$

Any clues will be appreciated.
Thanks

**TheEmptySet** · Jan 19th 2011, 12:13 PM

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

$<br /> \frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}<br />$

Any clues will be appreciated.
Thanks

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

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

Putting this into what you have above gives

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

can you finish from here?

**bugatti79** · Jan 19th 2011, 02:32 PM

Originally Posted by TheEmptySet

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

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

Putting this into what you have above gives

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

can you finish from here?

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

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

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

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

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

**TheEmptySet** · Jan 19th 2011, 02:34 PM

Originally Posted by bugatti79

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

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

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

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

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.

**bugatti79** · Jan 19th 2011, 02:37 PM

can I retract the double reply...I sent it by accident? I dont want to violate the forum rules?

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