# Thread: Differential equation change of variable

1. ## Differential equation change of variable

I'm having some trouble getting to the solution of the following differential equation:

Given that $\displaystyle y=1$ at $\displaystyle x=2$, use the substitution $\displaystyle v=3x-y-3$ to solve

$\displaystyle \left(3x-y-1\right)\frac{dy}{dx}=\left(3x-y+3\right)$

When I differentiate $\displaystyle v$ with respect to $\displaystyle x$, I eventually find $\displaystyle \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}$ (my suspicion is that my error is in this step), and when I perform the relevant substitutions and integrate I do not get to the correct answer of

$\displaystyle 2\ln|3x-y-3|=y-x+1+2\ln 2$

Any help on this would be greatly appreciated.

2. ## Re: Differential equation change of variable

Are we agree that

$\displaystyle v\ =\ 3x - y - 3\ \Rightarrow\ \frac{dv}{dx}\ =\ 3 - \frac{dy}{dx}$

and hence

$\displaystyle \frac{dy}{dx}\ =\ 3 - \frac{dv}{dx}$

... ?

That should give you a much easier ride than your

$\displaystyle \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}$

3. ## Re: Differential equation change of variable

Originally Posted by tom@ballooncalculus
Are we agree that

...and hence

$\displaystyle \frac{dy}{dx}\ =\ 3 - \frac{dv}{dx}$

That should give you a much easier ride than your

$\displaystyle \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}$
Yes, that's much easier! Thanks.