$\displaystyle y=ax^2+bx$

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.

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- Oct 23rd 2011, 12:08 AMCthul[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 AMsbhatnagarRe: [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 AMCthulRe: [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 AMsbhatnagarRe: [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 AMCthulRe: [Derivatives] Finding the solution
Ah, thanks.

- Oct 23rd 2011, 01:05 AMsbhatnagarRe: [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 AMAckbeetRe: [Derivatives] Finding the solution