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