I have the equation $y'' + 2y' + y = 3e^{-t}$ which I must first find a particular solution for via variation of parameters and then check the solution with the method of undetermined coefficients. The characteristic equation here is (r+1)^2 = 0, so we have a repeated root. I'm not sure where to go with finding a solution via variation of parameters. Tips would be greatly appreciated.
[Math] Differential Equations Variation of Parameters, Repeated Roots
ordinary differential equations
Best Answer
From the characteristic equation, we know that the solution to the homogeneous ODE is: $$ y_h(t) = c_1\underbrace{e^{-t}}_{y_1(t)} + c_2 \underbrace{te^{-t}}_{y_2(t)} $$ Now assume that the general solution has the form: $$ y(t) = u_1(t)y_1(t) + u_2(t)y_2(t) $$ Then by letting $g(t) = 3e^{-t}$, the method of variation of parameters requires that we solve the system: $$ \begin{bmatrix} y_1(t) & y_2(t) \\ y'_1(t) & y'_2(t) \end{bmatrix} \begin{bmatrix} u'_1(t) \\ u'_2(t) \end{bmatrix} = \begin{bmatrix} 0 \\ g(t) \end{bmatrix} $$ To this end, we form the corresponding augmented matrix and row-reduce: $$ \left[\begin{array}{cc|c} e^{-t} & te^{-t} & 0 \\ -e^{-t} & e^{-t} - te^{-t} & 3e^{-t} \\ \end{array}\right] \sim \left[\begin{array}{cc|c} e^{-t} & te^{-t} & 0 \\ 0 & e^{-t} & 3e^{-t} \\ \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & t & 0 \\ 0 & 1 & 3 \\ \end{array}\right] $$ Thus, we find that $u'_2(t) = 3$ and $u'_1(t) = -3t$ so that $u_1(t) = \tfrac{-3}{2}t^2 + C$ and $u_2(t) = 3t + D$. Hence, we conclude that the general solution is: \begin{align*} y(t) &= (\tfrac{-3}{2}t^2 + C)e^{-t} + (3t + D)te^{-t} \\ &= \underbrace{Ce^{-t} + Dte^{-t}}_{y_h(t)} + \underbrace{\tfrac{3}{2}t^2e^{-t}}_{y_p(t)} \\ \end{align*}