Answer to Question #137427 in Differential Equations for Raziya Sultana

Question

Solve the following equations using Wronskian’s method y ՜ ՜ -3y ՜ +2y = e^2t /(e^t +1).

Answer 1

$\displaystyle y'' - 3y' + 2y = \frac{e^{2t}}{e^{t} + 1} \\ \textsf{The solution to the above equation is}\\ y = y_c + y_p \\ \textsf{where}\, y_c \, \textsf{is the complementary factor and}\\ y_p\, \textsf{ is the particular integral}\\ \textsf{The auxiliary equation is} \,m^2 - 3m + 2 = 0 \\ \begin{aligned} (m - 1)(m - 2) &= 0 \\ \therefore m &= 1, 2 \end{aligned}\\ \therefore y_c = C_1e^t + C_2e^{2t} \\ \textsf{The Wronskian of the two solutions is}\\ W(t) = \begin{vmatrix} e^t & e^{2t} \\ \frac{\mathrm{d}}{\mathrm{d}t}(e^t) & \frac{\mathrm{d}}{\mathrm{d}t}(e^{2t}) \end{vmatrix} = \begin{vmatrix} e^t & e^{2t} \\ e^t & 2e^{2t} \end{vmatrix} = 2e^{3t} - e^{3t} = e^{3t} \\ \textsf{Our particular solution will be given by}\\ y_p = V_1(t)e^{t} +V_2(t)e^{2t}\\ \textsf{Where}\, V_1(t) = -\int \frac{r(t)e^{2t}}{W(t)}\, \mathrm{d}t \\ \begin{aligned} V_1(t) &= -\int \frac{e^{2t}}{e^t + 1} \cdot \frac{e^{2t}}{e^{3t}}\,\mathrm{d}t \\ &= -\int \frac{e^{t}}{e^t + 1} \,\mathrm{d}t = -\ln(e^t + 1) + C \end{aligned}\\ \begin{aligned} V_2(t) = \int \frac{r(t)e^{t}}{W(t)}\, \mathrm{d}t \\ &= \int \frac{e^{2t}}{e^t + 1} \cdot \frac{e^{t}}{e^{3t}}\,\mathrm{d}t\\ &= \int \frac{1}{e^t + 1} \,\mathrm{d}t = \ln\left(\frac{e^t}{e^t + 1}\right) + C \end{aligned} \\ \textsf{The constant terms of the integration can be}\\\textsf{ignored since we are trying to find a non-constant}\\\textsf{solution to the differential equation}\\ \begin{aligned} \therefore y &= y_c + y_p \\&= C_1e^t + C_2e^{2t} -\ln(e^t + 1)e^{t} +\ln\left(\frac{e^t}{e^t + 1}\right)e^{2t} \end{aligned}$