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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Show that the differential equation is homogeneous and solve $x\large\frac{dy}{dx}$$-y+x\sin\bigg(\large\frac{y}{x}\bigg)=$$0$

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  • A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y)$ is said to be homogenous if $F(x,y)$ is a homogenous function of degree zero.
  • To solve this type of equations substitute $y = vx$ and $\large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
The given equation can be written as $\large\frac{dy}{dx} =\frac{ [y - x\sin(y/x)]}{x}$
$F(x,y) =\large\frac{ [y - x\sin(y/x)]}{x}$
$F(kx.ky) = \large\frac{[ky - kx\sin (ky/kx)]}{kx}$$ = k^0.F(x,y)$
Hence this is an homogenous funtion of degree zero.
Step 2:
Using the information in the tool box, let us substitute for $y$ and $\large\frac{dy}{dx}$
$v+x\large\frac{dv}{dx }= \frac{[vx - x\sin v]}{x}$
$v + x\large\frac{dv}{dx }=$$ [v-\sin v]$
Cancelling $v$ on both sides we get
$x\large\frac{dv}{dx }= $$- \sin v$
Seperating the variables we get
$\large\frac{dv}{\sin v }= -\frac{dx}{x}$
Step 3:
Integrating on both sides we get,
$\int cosec v dv = - \int\large\frac{ dx}{x}$
$\log|cosec v - \cot v| = -\log x + \log C$
$cosec v - \cot v = \large\frac{C}{x}$
$(\large\frac{1}{\sin v}) - (\large\frac{\cos v}{\sin v}) =\frac{ C}{x}$
$\large\frac{[1-\cos v]}{\sin v} = \frac{C}{x}$
$1 -\cos v = \large\frac{C\sin v}{x}$
Step 4:
Writing $v = \large\frac{y}{x }$ we get,
$1 - \cos(\large\frac{y}{x}) = \large\frac{C\sin(\Large\frac{y}{x})}{x}$
$x[1-\cos(\large\frac{y}{x})] =$$ C\sin(\large\frac{y}{x})$
This is the required solution.
answered Aug 13, 2013 by sreemathi.v

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