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

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The forum discussion centers on solving the initial-value problem defined by the third-order differential equation y'''=(x^2)(e^x) with initial conditions y(0)=1, y'(0)=-2, and y''(0)=3. The user applied integration by parts to derive the second derivative y'' and subsequently the first derivative y'. The final expression for y was proposed as (x^2+6x-12)e^x+(5/2)x^2+4x+13. Verification of the solution against the initial conditions is essential to confirm its correctness.

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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

 
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Math10 said:

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.
 
Thank you.
 

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