- #1
Math10
- 301
- 0
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?