## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 5 years ago Solve the differential equation by separation of variables y ln x (dx/dy) = (y+1/x)^2

• This Question is Closed
1. anonymous

If this got posted twice sry my tablet works funny on this site

2. anonymous

Is it a case where you don't know how to separate it, or do the actual integration?

3. anonymous

$y \log x \frac{dy}{dy}=\left( \frac{y+1}{x} \right)^2 = \frac{(y+1)^2}{x^2} \rightarrow \frac{y}{(y+1)^2}dy=\frac{dx}{x^2 \log x}$

4. anonymous

The fraction in y can be decomposed:$\frac{y}{(y+1)^2}=\frac{1}{y+1}-\frac{1}{(y+1)^2}$

5. anonymous

Damn...I misread dx/dy as dy/dx...hang on...

6. anonymous

It's simpler:$y \log x \frac{dx}{dy}=\frac{(y+1)^2}{x^2} \rightarrow x^2 \log x dx = \frac{(y+1)^2}{y}dy$

7. anonymous

For the left-hand side, make a substitution, u = log(x), then$x=e^u$and also$dx = e^u du$The right-hand side becomes:$\int\limits_{}{}x^2 \log x dx = \int\limits_{}{}(e^{u})^2u.e^udu=\int\limits_{}{}u e^{3u}du$This can be solved using integration by parts a couple of times (selecting u as your 'u' and e^u as your 'dv') to give$\frac{e^{3u}}{3}(u-\frac{1}{3})+c=\frac{x^3}{3}(\log x - \frac{1}{3})+c$

8. anonymous

The right-hand side becomes:$\int\limits_{}{}\frac{(y+1)^2}{y}dy=\int\limits_{}{}\frac{y^2+2y+1}{y}dy=\int\limits_{}{}y + 2 + \frac{1}{y}dt=\frac{y^2}{2}+2y+\log y + c$The constants are not necessarily this same, even though I've used c twice. Your solution is then$\frac{y^2}{2}+2y+\log y = \frac{x^3}{3}(\log x - \frac{1}{3})+c$

9. anonymous

Leave it like this.

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy