jameshl2008
New member
- Joined
- Nov 15, 2013
- Messages
- 1
y''' - 3y'' + 3y' - y = e^x + 1
Textbook answer: y = [c1e^x + c2xe^x + c3x^2e^x] + {(1/6)(x^3e^x) - 1}
I understand that the solution is given by y = yh + yp where yh is brackets and yp is in curly brackets.
I am having trouble understanding how to get yp, especially with the added constant. I was able to get the coefficient (1/6) in my calculations but I also do not understand where the x^3 term comes from.
A detailed explanation would be very helpful to me for future reference.
Thanks!
Textbook answer: y = [c1e^x + c2xe^x + c3x^2e^x] + {(1/6)(x^3e^x) - 1}
I understand that the solution is given by y = yh + yp where yh is brackets and yp is in curly brackets.
I am having trouble understanding how to get yp, especially with the added constant. I was able to get the coefficient (1/6) in my calculations but I also do not understand where the x^3 term comes from.
A detailed explanation would be very helpful to me for future reference.
Thanks!