# Thread: Hard equation

1. ## Hard equation

Hello , I am trying to solve this equation for 2hours but I don't know how to do it...mabye someone can help me? I would be very gratefull

2. ## Re: Hard equation

Divide through by x, then substitute \displaystyle \begin{align*} u = y' \end{align*}. The equation becomes first order linear, and so can be solved with an integrating factor.

3. ## Re: Hard equation

u can also use Runge Kutta method of fourth order

4. ## Re: Hard equation

Yes, though I can't imagine why you would use a complicated numerical method when it is easy to solve for an exact solution.

What I would do is this: let u= y' so we get the first order equation $xu'- 2u= -\frac{2}{x^2}$. That is linear so we can separate it, first solving $xu'= 2u$ which in turn can be separated to $\frac{du}{u}= 2\frac{dx}{x}$. Integrating $ln(|u|)= 2ln|x|+ C$ which is equivalent to $u(x)= cx^2$.

To find a "particular solution", use "variation of parameters": seek a solution of the form $u(x)= v(x)x^2$. Then [tex]u'= v'x^2+ 2xv[tex] and the differential equation becomes $xu'- 2u= x^3v'+ 2x^2v- 2x^2v= x^3v'= -\frac{2}{x^2}$ so $v'= -\frac{2}{x^5}$.

5. ## Re: Hard equation

yeah thats right but as it is an initial value problem i thought we can apply runge kutta method