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

$\begin{array}{1 1}(A)\;\cos(\large\frac{y}{x}) = \log x +e \\(B)\;\cos(\large\frac{y}{2x}) = \log x +e \\(C)\;\cos(\large\frac{2y}{x})= \log x +e \\ (D)\sin(\large\frac{y}{x}) = \log x +e\end{array} $

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  • To solve homogenous 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 Aug 1, 2013 by sreemathi.v

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