# Thread: how do I solve this?

1. ## how do I solve this?

dy/dx = (x^2)+(8y^2)/(3xy)? I know that you have to name a parmater v that takes the value y/x and substitute it back in but I'm lost after that.

2. Originally Posted by lord12
dy/dx = (x^2)+(8y^2)/(3xy)? I know that you have to name a parmater v that takes the value y/x and substitute it back in but I'm lost after that.
What is it you are trying to do to this expression? Integrate it? Differentiate it?

3. Originally Posted by lord12
dy/dx = (x^2)+(8y^2)/(3xy)? I know that you have to name a parmater v that takes the value y/x and substitute it back in but I'm lost after that.

$\frac{dy}{dx} = \frac{x^2 + 8y^2}{3xy}$

You can do a number of things. First the equation is homogeneous so if you let

$v = \frac{y}{x}$ as you have said, then $y = x v$ then $y' = x v' + v$, substitute giving

$x v' + v = \frac{x^2 + 8 x^2 v^2}{3x^2 v}$

can x's from the right and solve for v' giving something that is separable. The equation is also Bernoulli.

4. Ahh. I see.

$\frac{dy}{dx} = \frac{x^2}{3xy}+\frac{8y^2}{3xy}$

Which gives:

$\frac{dy}{dx} = \frac{x}{3y}+\frac{8y}{3x}$

Giving:

$\frac{dy}{dx} = \frac{1}{3}(\frac{y}{x})^{-1}+\frac{8}{3}\frac{y}{x}$

$\text{Let} v = \frac{y}{x}$

$y = vx$

Hence

$\frac{dy}{dx} = v +x\frac{dv}{dx}$

Hence:

$v +x\frac{dv}{dx} = \frac{1}{3}(v)^{-1}+\frac{8}{3}v$

5. how do you write this in terms of x and y?

6. Originally Posted by lord12
how do you write this in terms of x and y?
First you solve it for v as a function of x. Then you replace v with y/x.