(20 pts) Find a general solution to y" - 2y' + y = 2e

^{t}/(1 + t^{2}).

Finding a solution takes several steps that I'll combine into two.

**Find and solve the characteristic equation.**r^{2}- 2r + 1 = 0 has two roots both equal to 1. So we have two independent solutions to the homogeneous equationy" - 2y' + y = 0: y _{1}= e^{t}and y_{2}= te^{t}and y_{c}= C_{1}e^{t}+ C_{2}te^{t}.

**Use Variation of Parameters**to find a particular solution to the nonhomogeneous equationy" - 2y' + y = 2e ^{t}/(1 + t^{2}).Let g(t) = 2e

^{t}/(1 + t^{2}). Theny _{p}= u_{1}(t)y_{1}+ u_{2}(t)y_{2},where u

_{1}(t) = - Integral(y_{2}g·dt/W(y_{1}, y_{2})), while u_{2}(t) = Integral(y_{1}g·dt/W(y_{1}, y_{2})). The WronskianW(y _{1}, y_{2}) = e^{2t}. So we getu _{1}(t) = - Integral(te^{t}2e^{t}dt/(1 + t^{2})/e^{2t}) = -Integral(2t·dt/(1 + t^{2})).Therefore we can take u On the other hand,_{1}(t) = -ln(1 + t^{2}).u _{2}(t) = Integral(e^{t}2e^{t}dt/(1 + t^{2})/e^{2t}) = Integral(2dt/(1 + t^{2})).Therefore we can take u

_{2}(t) = 2tan^{-1}(t).We now have a general solution:

### Answer

y(t) = e ^{t}(C_{1}+ C_{2}t) - e^{t}ln(1 + t^{2}) + 2te^{t}tan^{-1}(t).