Solve the initial-value problem?

Question

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

 

Answer: y=(x^2-6x+12)e^x+x^2/2-8x-11

 

Please show all your work step by step. Thanks.

Answer 1

y'''(x)= (x^2)*(e^x)

 

y''(x)= (x^2)*(e^x)-2x*e^x+2e^x+c

 

y''(0)=3

 

c+2=3.      c=1

 

y'(x)= (x^2)(e^x) -4x*e^x+6e^x+1+x+ d

 

y'(x)=-2

 

6+1+d=-2

 

d=-9

 

y(x)=[(x^2)*(e^x)-2x*e^x+2e^x+1]-4[x*e^x-e^x]+6e^x+x +(x^2)/2  -9x. +e

 

y(0)=1

 

2+1+4+6+e=1

 

e=-12

 

y(x)=(x^2)*(e^x)-6x*(e^x)+12e^x-8x+(x^2)/2-11