Y2-3y1+2y=e^x/1+e^x,, solution by varation of paramaters(wronskiam)

To solve via variation of parameters, you first need two linearly independent solutions. You can find them by solving the homogeneous part.



has characteristic equation



which has roots . So the characteristic solution is



with two linearly independent solutions, and .

Now when using variation of parameters, you're looking for solutions and where




with denoting the Wronskian of the characteristic solutions.

You have





and so the particular solution is




The general solution is then




where the term from gets absorbed into the same term from .

Ithink ghe answer to this question is b just trying to others