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Find the particular solution satisfying the given condition $2xy+y^2-2x^2\large\frac{dy}{dx}$$=0;\;y=\;2\;when\;x=\;1 This question has appeared in model paper 2012 Can you answer this question? 1 Answer 0 votes Toolbox: • To solve homogenous differential equations put y = vx and \large\frac{dy}{dx}$$ = 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.