1. [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.

2. Re: [Derivatives] Finding the solution

Are you really sure that this is a derivative question?
Is $\displaystyle y$ a function of x?

3. 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

4. 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.

Ah, thanks.

6. 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 $\displaystyle x$.

7. Re: [Derivatives] Finding the solution

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.