How to solve this ODE?

Sep 2011
392
7
Mumbai (Bombay),Maharashtra,India
Hello,
$y'+x\sqrt{y}=x^2$

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?
 
Dec 2011
2,313
914
St. Augustine, FL.
Are you sure you've presented the ODE correctly? You're treating it as if it is linear, when it's not, at least as given.
 
Sep 2011
392
7
Mumbai (Bombay),Maharashtra,India
Are you sure you've presented the ODE correctly? You're treating it as if it is linear, when it's not, at least as given.
Hello,
You mean to say the given equation is not in a standard form and my standard form is incorrect. So shall i square the whole euqtion? and try to solve it.
 
Dec 2011
2,313
914
St. Augustine, FL.
It's not a linear ODE. Best I can tell, it's some kind of Chiri's equation.
 
Sep 2011
392
7
Mumbai (Bombay),Maharashtra,India
It's not a linear ODE. Best I can tell, it's some kind of Chiri's equation.
Hello,
What is that? Would you explain something more about that?
 
Dec 2011
2,313
914
St. Augustine, FL.
Hello,
What is that? Would you explain something more about that?
It's a type of equation I know nothing about. We'll have to wait for someone more knowledgeable to come along and show us how to solve it. :)