Find the general solution of the given differential equation

[jameshl2008]

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!

 

[Deleted member 4993]

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!

First thing to do - did you check whether the given solution satisfies the the given ODE or not!?

Choice of particular solution - in my mind - is somewhat experience based (nice way to say arbitrary or really don't know the reason behind the choice , but looks plausible)

If I were to find y_p - given the ODE and the forcing function - I would choose first Axⁿe^x + B.

The constant is in y_p because we have "1" in forcing function.

Then use the ODE to solve for A, n & B.

 

[HallsofIvy]

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!

As long as the "right hand side", here \(\displaystyle e^x+ 1\), is of the form we expect to get for solutions of linear differential equations (exponentials, sine or cosine, polynomials and combinations of those) we can be sure that the "yp" must be of the same kind. So seeing \(\displaystyle e^x+ 1\), we should first think of \(\displaystyle Ae^x+ B\), "\(\displaystyle e^x\)" and "1" multiplied by as yet unknown constants. BUT, we already have \(\displaystyle e^x\), \(\displaystyle xe^x\), and \(\displaystyle x^2e^x\) as solutions to the associated homogeneous equation. There is NO constant that will give the right side- they will all give "0" and 0 times any constant is still 0. That's why we try the "next" possibility, \(\displaystyle x^3e^x\).

If we set \(\displaystyle y= Ax^3e^x+ B\) then \(\displaystyle y'= 3Ax^2e^x+ Ax^3e^x\), \(\displaystyle y''= 6Axe^x+ 6Ax^2e^x+ Ax^3e^x\), and \(\displaystyle y'''= 6Ae^x+ 18Axe^x+ 18Ax^2e^x+ 6Ax^3e^x\). We know have \(\displaystyle Ae^x\), \(\displaystyle 18Axe^x\), \(\displaystyle 18Ax^2e^x\), and \(\displaystyle 6Ax^3e^x\). If you put that into the differential equation, you will see that all except the \(\displaystyle Ae^x\) term cancel. That is because \(\displaystyle e^x\), \(\displaystyle xe^x\), and \(\displaystyle x^2e^x\) all satisfy the associated homogeneous equation- they all give "0".