Question

Solve the following initial value problem:-

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

Answer 1

We have,
$$\frac{dy}{dx}+y\tan x=2x+x^2\tan x.....\left(1\right)$$
Clearly, it is a linear differential equation of the form
$$\frac{dy}{dx}+Py=Q$$
$$\text{ where }P=\tan x\text{ and }Q=x^2\cot x+2x$$
$$\therefore I.F.=e^{\int Pdx}$$
$$=e^{\int \tan xdx}$$
$$=e^{log|\sec x|}=\sec x$$
$$\text{Multiplying both sides of }\left(1\right)\text{ by }I.F.=\sec x,\text{ we get }$$
$$\sec x(\frac{dy}{dx}+y\tan x)=\sec x(x^2\tan x+2x)$$
$$\Rightarrow \sec x(\frac{dy}{dx}+y\tan x)=x^2\tan x\sec x+2x\sec x$$
Integrating both sides with respect to x, we get

$$\Rightarrow y\sec x=\int x^2\tan x\sec xdx+2secx\int xdx-2\int [\frac{d}{dx}\left(secx\right)\int xdx]dx+C$$
$$\Rightarrow y\sec x=\int x^2\tan x\sec xdx+x^2\sec x-\int x^2\tan x\sec xdx+C$$
$$\Rightarrow y\sec x=x^2\sec x+C$$
$$\Rightarrow y=x^2+C\cos x.....\left(2\right)$$
Now, 
$$y\left(0\right)=1$$
$$\therefore 1=0+C\cos 0$$
$$\Rightarrow C=1$$
$$\text{Putting the value of C in }\left(2\right),\text{ we get }$$
$$y=x^2+\cos x$$
$$\text{ Hence, }y=x^2+\cos x\text{ is the required solution .}$$