Question

For each the differential equations given, find the general solution :
${\cos }^{2}x\frac{dy}{dx}+y=\tan x(0\le x\le \frac{\pi }{2})$

Answer 1

${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

$\frac{dy}{dx}+\left({\sec }^{2}x\right)y={\sec }^{2}x\tan x$

$\frac{dy}{dx}+Py=Q$

$P={\sec }^{2}x,Q={\sec }^{2}x\tan x$

$I.F.=e^{\int Pdx}$

$=e^{\int {\sec }^{2}xdx}$

$=e^{\tan x}$

$y\times I.F=\int Q\times I.Fdx+c$

$y\times e^{\tan x}=\int \left({\sec }^{2}x\tan x\right)\left(e^{\tan x}\right)dx+c$

$y{e}^{\tan x}=e^{\tan x}(\tan x-1)+c$

$y=e^{\tan x}+C$