Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}\text{ I.F. }& ={\mathrm{e}}^{\int 2{\sec }^{2}xdx}\\ & ={\mathrm{e}}^{2\tan x}\end{array}$
Solution of diff. eq.
$\begin{array}{c}y\cdot e^{2\tan x}=\int e^{2\tan x}\left(2{\sec }^{2}x+3\tan x\cdot {\sec }^{2}x\right)dx\\ y\cdot e^{2\tan x}=\int e^{2\tan x}\cdot \left(2{\sec }^{2}x\right)dx+\int e^{2\tan x}\cdot \left(3\tan x\cdot {\sec }^{2}x\right)dx\end{array}$
y. ${\mathrm{e}}^{2\tan x}={\mathrm{e}}^{2\tan x}\cdot 2\tan x-\int {\mathrm{e}}^{2\tan x}\cdot 2{\sec }^{2}x\times 2\tan xdx+\int {\mathrm{e}}^{2\tan x}\cdot 3\tan x\cdot {\sec }^{2}xdx$
$\begin{array}{rl}& y\cdot e^{2\tan x}=2\tan x\cdot e^{2\tan x}-\int {\mathrm{e}}^{2\tan x}\cdot \tan x{\sec }^{2}xdx\\ & y.e^{2\tan x}=2\tan x\cdot e^{2\tan x}-\frac{\tan x\cdot e^{2\tan x}}{2}+\frac{e^{2\tan x}}{4}+C\end{array}$
$\begin{array}{rl}& y=2\tan x-\frac{\tan x}{2}+\frac{1}{4}+{\mathrm{Ce}}^{-2\tan x}\\ & x=0,y=\frac{5}{4}\\ & c=1\end{array}$
$\mathrm{y}\left(\frac{\pi }{4}\right)=\frac{7}{4}+{\mathrm{e}}^{-2}$
Then $12\left(y\left(\frac{\pi }{4}\right)-{\mathrm{e}}^{-2}\right)=12\left(\frac{7}{4}\right)=21$