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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the differential equation : $\cos ^2 x \frac{d y}{d x}+y=\tan x$
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Answer

$ \cos ^2 x \frac{d y}{d x}+ y =\tan x$
$ \therefore \frac{d y}{d x}+\frac{y}{\cos ^2 x}=\frac{\tan x}{\cos ^2 x}$
$ \therefore \frac{d y}{d x}+\sec ^2 x \cdot y=\tan x \cdot \sec ^2 x $
The given equation is of the form
$\frac{d y}{d x}+P y= Q$
Where $P = sec^2 x$ and $Q = \tan x. sec^2 x$
$\therefore$ I.F. $=e^{\int P d x}=e^{\int \sec ^2 x d x}=e^{\tan x }$
$\therefore $ Solution of the given equation is
$ y (\text { I.F. })=\int Q \cdot(I . F .) d x+c$
$ \therefore ye ^{\tan x }=\int \tan x \cdot \sec ^2 x \cdot e^{\tan x} d x+c$
$ \text { Put \tan x } t$
$ \therefore \sec ^2 xdx = dt$
$ \therefore ye ^{\tan x }=\int t e^t d t+c$
$ =t \int e^t d t-\int\left[\frac{d}{d t}(t) \int e^t d t\right] d t+c$
$ =t e^t-\int e^t d t+c$
$ = te ^{ t }- e ^{ t }+ c$
$ \therefore ye ^{\tan x }= e ^{\tan x }(\tan x -1)+ c $
$\therefore y=\tan x-1+c \cdot e^{-\tan x}$

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