1. ## [SOLVED] Please solve this

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

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

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

2. Originally Posted by Altair
$(3x+y)dy = (x+3y)dx , y(0) = 1$

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

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

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

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

Substitute y = xv:

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

The DE is now seperable:

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

etc.

3. Oh sorry. Thanks.