# 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$

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

Toolbox:
• 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}) $$= \log x + log c Step 3: Given y = 0 and x = 1 we get, \cos(0) = \log(1) + \log C, since \cos 0 = 1 and \log 1 = 0 \log_e C= 1 C = e^1 C=e Substituting for C we get \cos(\large\frac{y}{x})$$= \log x +e$
This is the required equation.
answered Aug 1, 2013