Initial Value Problem

  • Thread starter alyston
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  • #1
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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.
 

Answers and Replies

  • #2
dextercioby
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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
 

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