- Sep 2011

- 392

- 7

- Mumbai (Bombay),Maharashtra,India

$y'+x\sqrt{y}=x^2$

Solution:- This ODE is already in standard form.Thus P(x)=x. And we have$\int P(x)dx=\int x dx=\frac{x^2}{2}$ So integrating factor is $e^{\int P(x)dx}=e^{\frac{x^2}{2}}$

Therefore,multiplying both sides of standard form by $e^{\frac{x^2}{2}}$ yields

$e^{\frac{x^2}{2}}y'+x*\sqrt{y}e^{\frac{x^2}{2}}=x^2*e^{\frac{x^2}{2}}$

$\frac{d[\sqrt{y}*e^{\frac{x^2}{2}}]}{dx}=x^2*e^{\frac{x^2}{2}}$

Now, integrating both sides we get

$\sqrt{y}*e^{\frac{x^2}{2}}=\int x^2*e^{\frac{x^2}{2}}$

How to proceed further?

I think integral of right side of this equation contains error function. How to solve that integral?