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Find the general solution. cos x^ 2 dy/dx + y = tan x ( 0 less than or equal to x < pi/2)

Find the general solution.

    Q5.    \cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )

Answers (1)

Given equation is
\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )
we can rewrite it as
\frac{dy}{dx}+\sec^2x y= \sec^2x\tan x
This is  \frac{dy}{dx} + py = Q  where p = \sec ^2x and Q =\sec^2x \tan x
I.F. = e^{\int pdx}= e^{\int \sec^2 xdx}= e^{\tan x}                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(e^{\tan x}) =\int ((\sec^2 x\tan x)\times e^{\tan x})dx +C
ye^{\tan x} =\int \sec^2 x\tan xe^{\tan x}dx+C\\
 e^{\tan x } = t\\ \Rightarrow \sec^2x.e^{\tan x}dx = dt
\int t.\log t dt = \log t.\int tdt-\int \left ( \frac{d(\log t)}{dt}.\int tdt \right )dt \\ \\ \int t.\log t dt = \log t . \frac{t^2}{2}- \int (\frac{1}{t}.\frac{t^2}{2})dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \int \frac{t}{2}dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \frac{t^2}{4}\\ \\ \int t.\log t dt = \frac{t^2}{4}(2\log t -1)
Now put again t = e^{\tan x}
\int \sec^2x\tan xe^{\tan x}dx = \frac{e^{2\tan x}}{4}(2\tan x-1)
Put this value in our equation

ye^{\tan x} =\frac{e^{2\tan x}}{4}(2\tan x-1)+C\\ \\
Therefore, the general solution is  y =\frac{e^{\tan x}}{4}(2\tan x-1)+Ce^{-\tan x }\\