general solution to differential equation

For our practice final one of the questions is

Find the general solution to the following differential equation: y'' - 4y' + 4y =x^{2 }- 3x + 2

I understand that one would add the particular solution and general solution. For the particular solution I got 1/4x^{2}-1/4x+1/8 and for the general solution I got Ae^{2x}+Be^{2x}. However he gave us the solution and it was Ae^{-2x}+Bxe^{-2x}+1/4x^{2}-1/4x+1/8. Does anyone see where I went wrong? I can show more work if thats helpful.

Re: general solution to differential equation

Quote:

Originally Posted by

**JML2618** For our practice final one of the questions is

Find the general solution to the following differential equation: y'' - 4y' + 4y =x^{2 }- 3x + 2

I understand that one would add the particular solution and general solution. For the particular solution I got 1/4x^{2}-1/4x+1/8 and for the general solution I got Ae^{2x}+Be^{2x}. However he gave us the solution and it was Ae^{-2x}+Bxe^{-2x}+1/4x^{2}-1/4x+1/8. Does anyone see where I went wrong? I can show more work if thats helpful.

Your mistake was in not dealing with the fact that your characteristic polynomial has a single root of order 2.

That makes your homogeneous solution of the form

$\displaystyle y[x]=c_1e^{2x}+c_2xe^{2x}$ (note the factor x in the 2nd term)

go use this form for your homogeneous solution and resolve your particular solution.

Re: general solution to differential equation

oh okay i understand now thanks for the reply! Do you use this whenever it's in the form y''+y'+y and ce^ax +ce^ax when it's y''+y'?

Re: general solution to differential equation

or is it that the roots are equal?

Re: general solution to differential equation

Quote:

Originally Posted by

**JML2618** or is it that the roots are equal?

Pauls Online Notes : Differential Equations - Repeated Roots

Re: general solution to differential equation

Notice that $\displaystyle Ae^{2x}+ Be^{2x}= Ce^{2x}$ with C= A+ B. Those are not two **independent** solutions and cannot give the general solution to a second order equation.