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Show that the differential equation is homogeneous and solve $\left\{x\cos\bigg(\large\frac{y}{x}\bigg)\normalsize+y\sin\bigg(\large\frac{y}{x}\bigg)\right\}ydx=\left\{y\sin\bigg(\large\frac{y}{x}\bigg)-\normalsize x\cos\bigg(\large\frac{y}{x}\bigg)\right\}\normalsize xdy$

$\begin{array}{1 1} C=\large\frac{1}{xy}\sec\big(\large\frac{y}{x}\big) \\ C=\large\frac{1}{xy}\tan \big(\large\frac{y}{x}\big) \\C=\large\frac{1}{xy}\sin \big(\large\frac{y}{x}\big) \\C=\large\frac{1}{xy}\cos \big(\large\frac{y}{x}\big) \end{array} $

1 Answer

  • 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 equation can be rearranged and written as $\large\frac{dy}{dx} = \large\frac{y[x\cos(y/x) + y\sin(y/x)]}{x[y\sin(y/x) - x\cos(y/x)]}$
$F(x,y) = \large\frac{y [x\cos(y/x) +y\sin(y/x)]}{x[(y\sin(y/x) - x\cos((y/x)]}$
$F(kx,ky) = \large\frac{ky[kx\cos(ky/kx) + kysin(ky/kx)]}{kx[ky\sin(ky/kx) - kxcos(ky/kx)] }=$$ k^0.F(x,y)$
Hence this is a homogenous equation with degree zero.
Step 2:
Using the information in the tool box, let us substitute for $y$ and $\large\frac{dy}{dx}$
$v + x\large\frac{dv}{dx }$$= x(\cos v + vx\sin v).vx/(vx\sin v - x\cos v).x$
cancelling $x$ throughout we get
$v + x\large\frac{dv}{dx} =\frac{ (v\cos v + v^2\sin v)}{(v\sin v - \cos v)}$
Bringing $v$ from LHS to the RHS we get
$\large\frac{(v\cos v + v^2\sin v - v^2\sin v + v\cos v)}{(vsinv - cosv)}$
$x\large\frac{dv}{dx }=$$ \large\frac{2v\cos v}{(v\sin v - \cos v)}$
seperating the variables we get,
$\large\frac{(v\sin v - \cos v)}{v\cos v }= \frac{2dx}{x}$
Step 3:
Integrating on both sides we get,
$\int\large\frac{ v\sin v}{v\cos v}$$dv - \int\large\frac{\cos v}{\cos v}$$dv = 2 \int\large\frac{dx}{x}$
$\int \tan v - \int (\large\frac{1}{v})$$dv = 2 \int\large\frac{ dx}{x}$
$\log(\sec v) - \log v = 2 \log x + \log C$
$\log(\large\frac{\sec v}{v}) = $$\log Cx^2$
$\large\frac{\sec v}{v}$$= Cx^2$
$\sec v = Cx^2v$
Step 4:
Writing $v = \large\frac{y}{x}$ we get,
$\sec(\large\frac{y}{x} )= $$Cx^2(\large\frac{y}{x} )$
This is the required equation.
answered Aug 13, 2013 by sreemathi.v

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