Thread: [SOLVED] Please solve this

$\displaystyle (3x+y)dy = (x+3y)dx , y(0) = 1$

As it is a homogeneous differential equation I substituted $\displaystyle y=vx$ and I can't solve it further than the following expression :-

$\displaystyle \frac{x}{dx} = \frac {(1+2v)}{(3+v)dv}$

Please simplify it for me.

Originally Posted by Altair

$\displaystyle (3x+y)dy = (x+3y)dx , y(0) = 1$

As it is a homogeneous differential equation I substituted $\displaystyle y=vx$ and I can't solve it further than the following expression :-

$\displaystyle \frac{x}{dx} = \frac {(1+2v)}{(3+v)dv}$

Please simplify it for me.

I'm sorry but what working out you've done is wrong.

$\displaystyle \frac{dy}{dx} = \frac{x + 3y}{3x + y} = \frac{1 + 3 \, (\frac{y}{x})}{3 + \, (\frac{y}{x})}$.

Substitute y = xv:

$\displaystyle v + x \frac{dv}{dx} = \frac{1 + 3v}{3 + v} \Rightarrow x \frac{dv}{dx} = \frac{1 + 3v}{3 + v} - v = \frac{1 - v^2}{3 + v}$.

The DE is now seperable:

$\displaystyle \frac{3 + v}{v^2 - 1} \, dv = \frac{1}{x} \, dx$

etc.

Oh sorry. Thanks.