Roger Smith

2021-12-10

I am trying to solve the following:
$y{}^{″}+4{y}^{\prime }=\mathrm{tan}\left(t\right)$
I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:
${u}_{1}=\int \frac{\mathrm{tan}t\mathrm{cos}2t}{2}$
I am stuck here

yotaniwc

$\mathrm{cos}2t=2{\mathrm{cos}}^{2}t-1$ hence
$\int \frac{\mathrm{tan}t\mathrm{cos}2t}{2}$$dt=\int \mathrm{sin}t\mathrm{cos}tdt-\frac{1}{2}\mathrm{tan}tdt$
$=\frac{1}{2}{\mathrm{sin}}^{2}t-\frac{1}{2}\mathrm{ln}\left(\mathrm{sec}t\right)+C$
$=\frac{1}{2}\left({\mathrm{sin}}^{2}t+\mathrm{ln}\left(\mathrm{cos}t\right)\right)+C$

SlabydouluS62

The homogeneous are ${y}_{1}\left(t\right)=\mathrm{sin}2t$ and $\mathrm{cos}2t$. In the method of variations, we form the particular solution ${y}_{p}\left(t\right)$​​​​​​​ as
${y}_{p}\left(t\right)={C}_{1}\left(t\right){y}_{1}\left(t\right)+{C}_{2}\left(t\right){y}_{2}\left(t\right)$
where the functions ${C}_{1}$ and ${C}_{2}$ are given by
${C}_{1}=-\int \frac{1}{W\left(t\right)}{y}_{2}\left(t\right)\mathrm{tan}tdt\phantom{\rule{0ex}{0ex}}{C}_{2}=+\int \frac{1}{W\left(t\right)}{y}_{1}\left(t\right)\mathrm{tan}tdt$
where $W\left(t\right)$ is the Wronskian for ${y}_{1}$ and ${y}_{2}$
First, the Wronskian is trivially evaluated to be $W=-2$
Second, we evaluate ${C}_{1}$ and ${C}_{2}$
${C}_{1}=\frac{1}{2}\int \mathrm{cos}2t\mathrm{tan}tdt\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \left(\mathrm{sin}2t-\mathrm{tan}t\right)dt$
$=-\frac{1}{4}\mathrm{cos}2t+\frac{12}{\mathrm{log}\left(\mathrm{cos}t\right)}$
${C}_{2}=-\frac{1}{2}\int \mathrm{sin}2t\mathrm{tan}tdt\phantom{\rule{0ex}{0ex}}=-\int {\mathrm{sin}}^{2}tdt\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}t+\frac{1}{4}\mathrm{sin}2t$
Third, we determine ${y}_{p}$ as
${y}_{p}\left(t\right)=\left(-\frac{14}{\mathrm{cos}2}t+\frac{12}{\mathrm{log}\left(\mathrm{cos}t\right)}\right)\mathrm{sin}t+\left(-\frac{12}{t}+\frac{14}{\mathrm{sin}2}t\right)\mathrm{cos}$
$=-\frac{12}{t}\mathrm{cos}2t+\frac{12}{\mathrm{sin}2}t\mathrm{log}\left(\mathrm{cos}t\right)$
Finally, the total solution to the ODE is
$y\left(t\right)=A\mathrm{sin}2t+B\mathrm{cos}2t-\frac{12}{t}\mathrm{cos}2t+\frac{12}{\mathrm{sin}2}t\mathrm{log}\left(\mathrm{cos}t\right)$

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