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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the particular solution satisfying the given condition $2xy+y^2-2x^2\large\frac{dy}{dx}$$=0;\;y=\;2\;when\;x=\;1$

This question has appeared in model paper 2012

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1 Answer

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Toolbox:
  • To solve homogenous differential equations put $y = vx$ and $\large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
Given : $2xy + y^2 -2x^2 \large\frac{dy}{dx}$$ = 0$
Rewriting this as $\large\frac{dy}{dx} = \frac{(y^2 + 2xy)}{2x^2}$
Using the information in the tool box
$v + x\large\frac{dv}{dx} =\frac{( v^2x^2 + 2xvx)}{2x^2}$
on taking $x^2$ as the common factor and cancelling we get,
$v + x\large\frac{dv}{dx} = \frac{(v^2+2v)}{2}$
Bringing v to the RHS
$\large\frac{xdv}{dx }= \frac{(v^2 +2v - 2v)}{2}$
$x\large\frac{dv}{dx} = \frac{v^2}{2}$
seperating the variables we get
$2\large\frac{dv}{v^2 }= \frac{dx}{x}$
Step 2:
Integrating on both sides we get
$2\large\frac{v^{-1}}{-1} =$$\log x + C$
$\large\frac{2}{v}$$=\log x+C$
Substituting for $v =\large\frac{y}{x}$
$\large\frac{-2x}{y }$$= \log x + C$
Step 3:
Given $x = 1$ and $y = 2$
$\large\frac{-2}{2 }=$$ \log 1 + C $. But $\log 1 = 0$
$C=-1$
Substituting the value for $C$ we get
$\large\frac{-2x}{y }$$=\log x - 1$
$-2x = y[ \log x - 1]$
$y[\log x -1] + 2x = 0$ is the required equation.
answered Aug 1, 2013 by sreemathi.v
 

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