Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Find the general solution of the differential equation $\cos^2x\large\frac{dy}{dx}$$+y=\tan x\bigg(0\leq x<\large\frac{\pi}{2}\bigg)$

Can you answer this question?

1 Answer

0 votes
  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i) Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dx + C$
Step 1:
The given equation is a first order linear equation
Let us write the equation by dividing throughout by $\cos^2x$
$\large\frac{dy}{dx} +\frac{ y}{\cos^2x }=\frac{ \tan x}{\cos^2x}$
$\large\frac{dy}{dx}$$ + (\sec^2x) y = \sec^2x \tan x $
Step 2:
Let us find the I.F
$\int Pdx = \int \sec^2xdx = \tan x$
Hence $I.F = e^{\large\tan x}$
Hence the solution is $y.e^{\tan x} = \int\sec^2x \tan x.e^{\tan x} + C$
To integrate this we will have to follow the substitution method.
Let $\tan x = t$; then $dt = \sec^2dx$
$y.e^{\tan x} = \int t.e^t. dt + C$
This has to be solved by integration by parts.
Step 3:
Let $u = t$ and $du = dt$ and $dv = e^tdt$, hence $v = e^t$
$\int t.e^tdt = (t.e^t) - \int e^tdt$
$\qquad\;\;\;\;= t.e^t - e^t + C$
Substituting the value of t we get,
Hence the required equation is $y.e^{\tan x }= \tan x.e^{\tan x} - e^{\tan x }+ C$
$y.e^{\tan x} = e^{\tan x}(\tan x - 1) + C$
$y = (\tan x -1) +\large\frac{ C}{e^{\Large\tan x}}$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App