# Thread: [SOLVED] Differentiation in explicit form

1. ## [SOLVED] Differentiation in explicit form

Find the general solution of the differential equation

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

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

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

$y = xu$

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

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

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

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

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

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

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:

$u = \text{ something }$

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

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

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

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

$y = xu$

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

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

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

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

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

$\frac{du}{u} = 2 \times \frac{dx}{x}$
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?
Read post #4 more carefully.

7. Originally Posted by ronaldo_07
This is all a bit confusing to me. To find y do you not need to integrate somewhere?
At the very end of my first post you must apply integrated with respect to u on the LHS, and with respect to x on the RHS. Then go to my 2nd post which tells you how to proceed to change back from u to y.