Ask Questions, Get Answers

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

Find the particular solution, satisfying the given condition, for the following differential equation : $\large \frac{dy}{dx}-\frac{y}{x}$$+cosec \bigg( \large\frac{y}{x} \bigg)$$=0; y=0\: when \: x = 1$

Can you answer this question?

1 Answer

0 votes
  • To solve homogeneous differential equation, put $y = vx$ and $\large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
Let us rearrange and write the equation as $\large\frac{dy}{dx} = (\large\frac{y}{x})$$ - cosec(\large\frac{y}{x})$
Using the information in the tool box,
$v+x\large\frac{dv}{dx}$$ = v + cosec v $
cancelling $v$ on both sides we get,
$x\large\frac{dv}{dx }$$= - cosec v$
Seperating the variables we get,
$\large\frac{dv}{cosec v }= \frac{- dx}{x}$
Since $\large\frac{1}{cosec v}$$ = \sin v$
$\sin v dv = -\large\frac{ dx}{x}$
Step 2:
Integrating on both sides we get,
$\int \sin vdv = - \int\large\frac{dx}{x}$
$- \cos v = -\log x - C$
$ \cos v = \log x + C$
Substituting for $v$ we get,
$\cos(\large\frac{y}{x}) $$= \log x + log c$
Step 3:
Given $y = 0$ and $x = 1$ we get,
$\cos(0) = \log(1) + \log C$, since $\cos 0 = 1$ and $\log 1 = 0$
$\log_e C= 1$
$C = e^1$
Substituting for $C$ we get
$\cos(\large\frac{y}{x}) $$= \log x +e$
This is the required equation.
answered Sep 23, 2013 by sreemathi.v

Related questions

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