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.

 

[dextercioby]

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