Answer to Question #137084 in Differential Equations for shanto

Question

Solve the following equations using Wronskian’s method:
y ՜ ՜-3y ՜+2y = e2t /(et +1).

Answer 1

Let us consider the method from this book (http://scipp.ucsc.edu/~haber/ph116C/Wronskian_12.pdf page 4)

Let the general solution be

$y(t) =c_1y_1(t) +c_2y_2(t) +y_p(t),$ $y_1, y_2$ are the solutions of the homogeneous equation. These solutions may be obtained by the characteristic equation $\lambda^2 -3\lambda + 2 = 0 , \;\;\; \lambda_1 = 1, \lambda_2 = 2 \; \Rightarrow \; y_1 = e^t, \; y_2 = e^{2t}$ .

Let us calculate the Wronskian:

$W = \det \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} = y_1y_2'-y_2y_1' = e^t\cdot2e^{2t} - e^{2t}\cdot e^t = e^{3t}.$

$y_p(t) = -y_1(t)\int\dfrac{y_2(t)}{W(t)}f(t)\,dt + y_2(t)\int\dfrac{y_1(t)}{W(t)}f(t)\,dt = \\ = -e^t\int\dfrac{e^{2t}}{e^{3t}}\cdot\dfrac{e^{2t}}{e^t+1}\,dt + e^{2t}\int\dfrac{e^{t}}{e^{3t}}\cdot\dfrac{e^{2t}}{e^t+1}\,dt = -e^t\int \dfrac{e^t}{e^t+1}\,dt + e^{2t}\int \dfrac{1}{e^t+1}\,dt = -e^t\ln(e^t+1) + e^{2t}(t-\ln(e^t+1)) = te^{2t} - \ln(e^t+1)\cdot(e^t+e^{2t}).$

Therefore, the solution is

$y(t) =c_1y_1(t) +c_2y_2(t) +y_p(t) = c_1e^t+c_2e^{2t} + te^{2t} - \ln(e^t+1)\cdot(e^t+e^{2t}).$