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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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True-or-False: Solution of $x\frac{dy}{dx} - y = x\tan\frac{y}{x}$ is $\sin\bigg(\frac{y}{x}\bigg)=kx$.

$\begin{array}{1 1}True \\ False \\ Cannot\;be\;determined \end{array}$

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  • To solve a homogenous linear differential equation put $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large \frac{dv}{dx}$
  • $\int \cot x dx =\log |\sin x|+c$
Divide throughout by $x$
$\large\frac{dy}{dx}-\frac{y}{x}$$=\tan \large\frac{y}{x}$
Put $y=vx$ hence on differentiating w.r.t x we get,
$v+x \large \frac{dv}{dx}=$$\large\frac{dy}{dx}$
On substituting this we get,
$v+x \large \frac{dv}{dx}$$-v=\tan v$
$=> x \large\frac{dv}{dx}$$=\tan v$
Now seperating the variables we get,
$\large\frac{dv}{\tan v}=\frac{dx}{x}$
$=>\cot v dv=\large\frac{dx}{x}$
Integrating on both sides we get,
$=>\int \cot v dv=\int \large\frac{dx}{x}$
$=>\log |\sin v|=\log x+\log c$
$=>\log |\sin v|=\log x+\log xk$
or $\sin v=kx$
Substituting for $v=y/x$
$\sin (y/x)=kv$
Therefore Solution of $x\large\frac{dy}{dx}$$-y=x\tan\large\frac{y}{x}$ is $\sin\bigg(\large\frac{y}{x}\bigg)$$=kx.$ is $True$
answered May 23, 2013 by meena.p

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