# Initial Value Problem

• alyston

## Homework Statement

d2y/dx2 = 2-6x

Given: y(0)=-3 and y'(0)=4

## Homework Equations

None that I know of.

## The Attempt at a Solution

I know that for a single order derivative you would just find the integral, set y=1 and x=0. But I'm confused because here we're given two conditions, instead of just one. And I don't know how to do this type of problem with a second order derivative. But I know that:

The second integral of 2-6x is x2-x3. So,
y=x2-x3 + C
1= C

I also know that the first integral is 2x-3x2

But where does y'(0)=4 come into place? Do I need two equations here?
Mainly, I need to find the value of C, which stands for Constant.

Thanks! I'd really appreciate any help.

If you integrate twice, each time you have to introduce an integration constant. That's why you need two additional conditions to find these constants.

So the antiderivative of the antiderivative of 2-6x is

a) x^2 - x^3.
b) x^2 - x^3 + C
c) x^2 - x^3 + Cx +D
d) x^2 - x^3 + C + D
e) all 4 of the above
f) none of the above 5 options