Solving the Initial-Value Problem y'''=(x^2)(e^x), y(0)=1, y'(0)=-2, y(0)=3

[Math10]

Homework Statement

Solve the initial-value problem y'''=(x^2)(e^x), y(0)=1, y'(0)=-2, y"(0)=3.

Homework Equations

Here's the work:

I did integration by parts and got y"=(x^2)(e^x)+(2x)(e^x)-2e^x+C
C=5
y"=(x^2)(e^x)+2x(e^x)-2e^x+5
y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+C
C=4
y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+4
and I did the same thing to get y. But I got y=(x^2+6x-12)e^x+(5/2)x^2+4x+13 as the answer. Is this answer right?

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

Solve the initial-value problem y'''=(x^2)(e^x), y(0)=1, y'(0)=-2, y"(0)=3.

Homework Equations

Here's the work:

I did integration by parts and got y"=(x^2)(e^x)+(2x)(e^x)-2e^x+C
C=5
y"=(x^2)(e^x)+2x(e^x)-2e^x+5
y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+C
C=4
y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+4
and I did the same thing to get y. But I got y=(x^2+6x-12)e^x+(5/2)x^2+4x+13 as the answer. Is this answer right?

I don't know, but you can easily check it yourself. Does your $y'''=x^2e^x$? Is $y(0)=1$? Is $y'(0) = -2$? Is $y''(0)=3$? If so, you have your answer.

 

[Math10]

Thank you.