Question

Solve $(1+y){\tan }^{2}x+\tan x\left(\frac{dy}{dx}\right)+y=0$

Answer 1

Simplify and Integrate the equation:

Given that,

$$\begin{gathered} (1+y){\tan }^{2}x+\tan x\left(\frac{dy}{dx}\right)+y=0 \\ \Rightarrow \tan x\left(\frac{dy}{dx}\right)+\left({\tan }^{2}x+1\right)y=-{\tan }^{2}x\left[\because \mathrm{Taking}y\mathrm{as}\mathrm{common}\mathrm{term}\right] \\ \Rightarrow \frac{\tan x}{\tan x}\left(\frac{dy}{dx}\right)+\left(\frac{{\tan }^{2}x}{\tan x}+\frac{1}{\tan x}\right)y=\frac{-{\tan }^{2}x}{\tan x}\left[\because \mathrm{Dividng}\mathrm{the}\mathrm{equation}\mathrm{by}\tan x\right] \\ \Rightarrow \frac{dy}{dx}+\left(\tan x+\cot x\right)y=-\tan x\left[\because \frac{1}{\tan x}=\cot x\right] \end{gathered}$$

Integrating the equation

$$\begin{gathered} I.F.=e^{\int \left(\tan \left(x\right)+\cot \left(x\right)\right)dx} \\ =e^{\ln \left(\sec \left(x\right)\right)+\ln \left(\sin \left(x\right)\right)} \\ =e^{\ln \left(\sec \left(x\right)·\sin \left(x\right)\right)} \\ =\tan x \end{gathered}$$

Now,

$$\begin{gathered} y\tan x=-\int {\tan }^{2}xdx \\ y\tan x=-\int (se{c}^{2}x-1)dx\left[\because {\tan }^{2}x=se{c}^{2}x-1\right] \\ y\tan x=-\tan x+x+c \end{gathered}$$

Therefore, the solution of $(1+y){\tan }^{2}x+\tan x\left(\frac{dy}{dx}\right)+y=0$ is $-\tan x+x+c$.