logo

Ask Questions, Get Answers

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

Find the particular solution satisfying the given condition $(x+y)dy+(x-y)dx=0;\;y=1;\;when\;x=\;1$

$\begin{array}{1 1} (A)\;2\tan^{-1}(\large\frac{y}{x}) +\log[x^2+y^2] =\large\frac{ \pi}{2} + \log 2. \\(B)\;2\tan^{-1}(\large\frac{y}{x}) - \log[x^2+y^2] =\large\frac{ \pi}{2} + \log 3. \\ (C)\;2\tan^{-1}(\large\frac{y}{x}) + \log[x^2-y^2] =\large\frac{ \pi}{2} + \log 2. \\ (D)\;2\tan^{-1}(\large\frac{y}{x}) + \log[x+y^2] =\large\frac{ \pi}{2} + \log 2.\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • To solve homogenous differnetial equation put $y = vx$ and $\large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
Let us rearrange and write the equation as $(x+y)dy = -(x-y)dx$
$\large\frac{dy}{dx }= -\frac{(x-y)}{(x+y)}$
Using the information in the tool box we get,
$v+x\large\frac{dv}{dx }= \frac{-(x-vx)}{x+vx) }$
Taking the common factor x and cancelling on the RHS we get
$v+x\large\frac{dv}{dx} = \frac{(1-v)}{(1+v)}$
Bringing v from LHS to the RHS we get,
$x\large\frac{dv}{dx }= \frac{(1-v-v+v^2)}{(1+v)}$
$x\large\frac{dv}{dx }=\frac{ (v^2+1)}{(1+v)}$
Seperating the variables we get,
$\large\frac{(1+v)dv}{(v^2+1) }= -\frac{ dx}{x}$
Step 2:
Integrating on both sides
$\int\large\frac{(1+v)dv}{(v^2+1) }= -\int\frac{ dx}{x}$
$\int\large\frac{1}{(v^2+1) }+ \int\large\frac{ v}{(v^2+1) }=- \int\large\frac{ dx}{x}$
$\tan^-1(v) + \large\frac{1}{2}$$[\log(1+v^2)] = - \log x + C$
Substituting $v = \large\frac{y}{x}$ we get,
$\tan^-1(\large\frac{y}{x}) +\large\frac{1}{2}$$[\log(1+(y^2/x^2)] = - \log x + C$
$2\tan^{-1}(\large\frac{y}{x}) + [\large\frac{\log(x^2 + y^2)}{x^2}] = $$- 2\log x + C$
$2\tan^-1(\large\frac{y}{x}) +\large\frac{\log[x^2 + y^2]}{x^2} . $$ x^2= C$
$2\tan^-1(\large\frac{y}{x} )$$+ \log[x^2+y^2] = C$
Step 3:
Given $x = 1$ and $y = 1$, by substituting we get
$2\tan^{-1}(1) + log[1+1] = C$
But $\tan^-1(1) =\large\frac{ \pi}{4}$
$2.(\large\frac{\pi}{4}) $$+ \log 2 = C$
$C = \large\frac{\pi}{2}$$ + \log 2$
Substituting for $C$ we get,
$2\tan^{-1}(\large\frac{y}{x}) $$+ \log[x^2+y^2]$$ =\large\frac{ \pi}{2} $$+\log2.$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...