How to Solve for Initial Value in an Integration Problem?

[manal950]

Homework Statement

Solve for initial Value

The attempt at a solution

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[HallsofIvy]

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.