# Thread: [SOLVED] Differentiation in explicit form

1. ## [SOLVED] Differentiation in explicit form

Find the general solution of the differential equation

y′ =$\displaystyle \frac{2x^2+y^2}{xy}$. in explicit form.

2. Originally Posted by ronaldo_07
Find the general solution of the differential equation

y′ =$\displaystyle \frac{2x^2+y^2}{xy}$. in explicit form.
Use a substitution of $\displaystyle u = \frac{y}{x}$

$\displaystyle y = xu$

$\displaystyle \frac{dy}{dx} = u + x\frac{du}{dx}$

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

$\displaystyle \frac{dy}{dx} = \frac{2x}{y}+\frac{y}{x}$

$\displaystyle u + x\frac{du}{dx} = \frac{2}{u}+u$

$\displaystyle x\frac{du}{dx} = \frac{2}{u}$

$\displaystyle \frac{du}{u} = 2 \times \frac{dx}{x}$

3. What are we solving for in this question?

4. Originally Posted by ronaldo_07
What are we solving for in this question?
Solve for u.

Get it in the form:

$\displaystyle u = \text{ something }$

Then substitute back in for y. You know that $\displaystyle u = \frac{y}{x}$. Hence:

$\displaystyle \frac{y}{x} = \text{ something }$

$\displaystyle y = \text{ something } \times x$

5. Originally Posted by Mush
Use a substitution of $\displaystyle u = \frac{y}{x}$

$\displaystyle y = xu$

$\displaystyle \frac{dy}{dx} = u + x\frac{du}{dx}$

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

$\displaystyle \frac{dy}{dx} = \frac{2x}{y}+\frac{y}{x}$

$\displaystyle u + x\frac{du}{dx} = \frac{2}{u}+u$

$\displaystyle x\frac{du}{dx} = \frac{2}{u}$

$\displaystyle \frac{du}{u} = 2 \times \frac{dx}{x}$
This is all a bit confusing to me. To find y do you not need to integrate somewhere?

6. Originally Posted by ronaldo_07
This is all a bit confusing to me. To find y do you not need to integrate somewhere?