# Solve the Differential equations y'' + 4y = 4 tan^{2} x

Solve the Differential equations $y+4y=4{\mathrm{tan}}^{2} x$
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hajavaF
We need to solve following differential equation
$y"+4y=4{\mathrm{tan}}^{2} x$
Let us first solve the homogeneous part.
$y"+4y=0$
Let $y={e}^{m}x$ be the trial solution. Then we get the auxiliary equation
${m}^{2}+4=0$
This gives us
$m=±2i$
Thus the solution of the homogeneous part of the differential equation be
${y}_{h}\left(x\right)={c}_{1}\mathrm{cos}\left(2x\right)+{c}_{2}\mathrm{sin}\left(2x\right)$
We use variation of parameter to calculate a
particular solution y_h. Let us take
${y}_{1}=\mathrm{cos}2x,{y}_{2}=\mathrm{sin}2x$
Then we get
$y{‘}_{1}=-2\mathrm{sin}2x,y{‘}_{2}=2\mathrm{cos}2x$
The wro
ian of is
$W={y}_{1}y{‘}_{2}-y{‘}_{1}{y}_{2}$
$=\left(\mathrm{cos}2x\right)\left(2\mathrm{cos}2x\right)-\left(-2\mathrm{sin}2x\right)\left(\mathrm{sin}2x\right)=2\ne 0$
In this case it is given that
$f\left(x\right)=4\mathrm{tan}\left(2x\right)$
Then by the formula we obtain
${u}_{1}=\int \frac{-{y}_{2}f}{W}dx$
$=\int \frac{-\left(\mathrm{sin}2x\right)\left(4\mathrm{tan}2x\right)}{2}dx$
$=-2\int \frac{{\mathrm{sin}}^{2}\left(2x\right)}{\mathrm{cos}\left(2x\right)}dx$
$=-2\int \left(\frac{1-{\mathrm{cos}}^{2}2x}{\mathrm{cos}2x}\right)dx$
$=2\int \left(\mathrm{cos}2x-\mathrm{sec}2x\right)dx$
$=2\int \left(\mathrm{cos}2x-\mathrm{sec}2x\right)dx$
$=\mathrm{sin}2x-\mathrm{ln}|\mathrm{sec}2x+\mathrm{tan}2x|$
Similarly
${u}_{2}=\int \frac{-{y}_{1}f}{W}dx$
$=\int \frac{\left(\mathrm{cos}2x\right)\left(4\mathrm{tan}2x\right)}{2}dx$
$=2\int \mathrm{sin}\left(2x\right)dx$
$=-\mathrm{cos}\left(2x\right)$
Thus by variation of parameters the patricular solution is
${y}_{p}={u}_{1}{y}_{1}+{u}_{2}{y}_{2}$
$=\left(\mathrm{sin}2x-\mathrm{ln}|\mathrm{sec}2x+\mathrm{tan}2x|\right)\left(\mathrm{cos}2x\right)+\left(-\mathrm{cos}2x\right)\left(\mathrm{sin}2x\right)$
${y}_{p}=-\mathrm{cos}2x\mathrm{ln}|\mathrm{sec}2x+\mathrm{tan}2x|$
Hence the general solution is
$y={y}_{c}+{y}_{p}={c}_{1}\mathrm{cos}2x+{c}_{2}\mathrm{sin}2x-\mathrm{cos}2x\mathrm{ln}|\mathrm{sec}2x+\mathrm{tan}2x|$