# [Derivatives] Finding the solution

• Oct 23rd 2011, 12:08 AM
Cthul
[Derivatives] Finding the solution
$\displaystyle y=ax^2+bx$

Find the solution for the constants a & b.

$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+36x=0$

I got as far as:

$\displaystyle y'=2ax+b$
$\displaystyle y''=2a$

And don't know what to do next.
• Oct 23rd 2011, 12:45 AM
sbhatnagar
Re: [Derivatives] Finding the solution
Are you really sure that this is a derivative question?
Is $\displaystyle y$ a function of x?
• Oct 23rd 2011, 12:51 AM
Cthul
Re: [Derivatives] Finding the solution
I think it might be a mistake or I'm supposed to solve for x. I still don't get the answer.
Apparently a=-3 and b=1
• Oct 23rd 2011, 12:53 AM
sbhatnagar
Re: [Derivatives] Finding the solution
If you were supposed to solve for $\displaystyle x$ then $\displaystyle x=\frac{-b \pm \sqrt{b^2+4ay}}{2a}$ (using quadratic formula), else the question doesn't make any sense.

If you were supposed to find $\displaystyle a$ and $\displaystyle b$ ,there would be 2 equations.
• Oct 23rd 2011, 01:00 AM
Cthul
Re: [Derivatives] Finding the solution
Ah, thanks.
• Oct 23rd 2011, 01:05 AM
sbhatnagar
Re: [Derivatives] Finding the solution
... also we can differentiate both L.H.S and R.H.S if and only if $\displaystyle y$ is a function of $\displaystyle x$.
• Oct 24th 2011, 03:32 AM
Ackbeet
Re: [Derivatives] Finding the solution
Quote:

Originally Posted by Cthul
$\displaystyle y=ax^2+bx$

Find the solution for the constants a & b.

$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+36x=0$

I got as far as:

$\displaystyle y'=2ax+b$
$\displaystyle y''=2a$

And don't know what to do next.

Plug the original function into the differential equation. Then equate like powers of x.