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

Last updated date: 25th Mar 2023

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

Given differential equation is $\dfrac{{dy}}{{dx}} + y\tan x = {\cos ^3}x$

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

$\

\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 \\

\ $

$ \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 = \dfrac{{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.

Given differential equation is $\dfrac{{dy}}{{dx}} + y\tan x = {\cos ^3}x$

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

$\

\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 \\

\ $

$ \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 = \dfrac{{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.

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