# Solve for initial Value

1. Nov 2, 2012

### manal950

1. The problem statement, all variables and given/known data
Solve for initial Value

The attempt at a solution

http://store2.up-00.com/Sep12/RjY55784.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. Nov 2, 2012

### HallsofIvy

Staff Emeritus
The way you have written this is wrong and very confusing! You have the initial value problem
$$\frac{d^2y}{dx^2}= 2- 6x$$
y'(0)= 4, y(0)= 1.

Since the right side is a function of x only this really just an interation problem.
Integrating once,
$$\frac{dy}{dx}= 2x- 3x^2+ C$$
You have "$d^2y/dx^2$" still on the left side which is "wrong and confusing".
Because dy/dx= 4 when x= 0, C= 4 so
$$\frac{dy}{dx}= 2x- 3x^2+ 4$$
which is what you have on your sixth line although now you have it equal to "y", rather than dy/dx.

Integrating again, $y= x^2- x^3+ 4x+ C$ and, since y(0)= 1, C= 1 (in your hand written derivation, that "1" looks an awful lot like a "2") so your final answer, $y(x)= x^2- x^3+ 4x+ 1$ is correct.