Question

Solve the differential equation ${\displaystyle \frac{dy}{dx}}+y\tan x={\cos }^{3}x$

Answer 1

Hint:-Use the integrating factor method to get the solution for the above problem .

Given differential equation is ${\displaystyle \frac{dy}{dx}}+y\tan x={\cos }^{3}x$
Let $\tan x=p,{\cos }^{3}x=q$
Let the integrating factor (I.F) = $e^{\int pdx}$
We know that
                 $p=\tan x$
Substitute the p value in I.F
$\Rightarrow e^{\int \tan xdx}$
$\Rightarrow e^{\ln (\sec x)}$ [$\because \int \tan xdx=\ln (\sec x)$]
$\Rightarrow \sec x$ [$\because$ $e$ is the inverse function of ln where it gets cancel]

Here the solution of equation is of the form:
$y(I.F)=\int q\times I.Fdx$
Now let us simplify the equation by substituting the values
$$\begin{gathered} \Rightarrow y.\sec x=\int {\cos }^{3}x\sec xdx \\ \Rightarrow y.\sec x=\int {\cos }^{3}x\left(\frac{1}{\cos x}\right)dx \end{gathered}$$
$\Rightarrow y.\sec x=\int {\cos }^{2}xdx$
$\Rightarrow y.\sec x=\int \frac{1+\cos 2x}{2}dx$ $[\because \cos 2x=2{\cos }^{2}x-1]$
$\Rightarrow y={\displaystyle \frac{x.\cos x}{2}}+\frac{1}{4}\sin 2x.\cos x+\cos x+C$
NOTE: In this kind of problems everyone solves the problems without using the integrating factor method (I.F) which is very important to use.