Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\cos }^{2}x\frac{dy}{dx}+y=\tan x \\ \Rightarrow \frac{dy}{dx}+\left({\sec }^{2}x\right)y=\left(\tan x\right){\sec }^{2}x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Px=Q,\mathrm{we}\mathrm{get} \\ P={\sec }^{2}x \\ Q=\left(\tan x\right)\left({\sec }^{2}x\right) \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int {\sec }^{2}xdx}=e^{\tan x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times e^{\tan x}=\int \left(\tan x\right)\left({\sec }^{2}x\right)\times e^{\tan x}dx+C \\ \Rightarrow y{e}^{\tan x}=I+C.....\left(1\right) \\ \mathrm{Now}, \\ I=\int \left(\tan x\right)\left({\sec }^{2}x\right)\times e^{\tan x}dx \\ \mathrm{Putting}t=\tan x,\mathrm{we}\mathrm{get} \\ dt={\sec }^{2}xdx \\ \therefore I=\int \underset{\mathrm{I}}{t}\times \underset{\mathrm{II}}{{\mathrm{e}}^t}dt \\ =t\times \int e^{t}dt-\int \left(\frac{dt}{dt}\times \int e^{t}dt\right)dt \\ =t{e}^t-\int e^{t}dt \\ =t{e}^t-e^t \\ \Rightarrow I=\tan x{e}^{\tan x}-e^{\tan x}=e^{\tan x}\left(\tan x-1\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}I\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y{e}^{\tan x}=e^{\tan x}\left(\tan x-1\right)+C \end{gathered}$$