Differential Equation: 2nd Order DFQ. Please Help

Question

Using the method of undetermined coefficients, determine the general solution of the following second-order, linear, non-homogenous equation.

 

y^''- 4y^' + 4y = 2e^2x+3

 

note: I assume Yp = Ae^2x+3, when I derived twice to get 2nd derivative and plug it in the original equation, I get 0 = 2e^2x+3. I couldn't solve for A. I don't know what to do from here.

Answer 1

The characteristic equation of the homogeneous equation y^''-4y^'+4y =0 is 

k²-4k+4=0 and k=2 is its repeated (twice) root and as noticed by Andrew G.,  it appears in the RHS of the inhomogeneous equation. This means that you have to "increase" the power of the polynomial in  the RHS, such that to be able to find a particular solution.

If you  search solution in the form y_p(x)=Ax²e^2x+3,  after substitution you get  (A-1)e^2x=0, -->A=1, So, the particular solution is y_p(x)=x²e^2x+3

A general solution is

y(x)=C₁e^2x+C₂e^2x+x²e^2x+3.

 

Answer 2

Notice that Yp(x) = A*exp (2x+3) = C*exp(2x) is a solution to the homogeneous differential equation, and that is why it equals zero when you plugged into the DEQ.  When the right side of the differential equation ( g(x) ) includes part of the complimentary solution, you must multiply the "guess" for Yp by x (or x^2 in this case because the characteristic equation has a double root of 2).  Try using Yp = A*x^2*exp(2x)

 

BTW, you can use Yp = A*x^2*exp(2x+3) if you want, it will only change the value of A.