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Show that the differential equation is homogeneous and solve $(x^2-y^2)dx+2xy\;dy=0$

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:
By rearranging the equation we get,
$\large\frac{dy}{dx} = -\large\frac{(x^2-y^2)}{2xy }$
Using the information in the tool box let us substitute $y = vx$ and $\large\frac{dy}{x} =$$ v + x\large\frac{dv}{dx}$
$v + x\large\frac{dv}{dx} = -\large\frac{(x^2-v^2x^2)}{2xvx}$
Step 2:
Taking $x^2$ as the common factor and cancelling it on the RHS
$v + x\large\frac{dv}{dx} = \frac{(1-v^2)}{2v}$
Bringing $v$ from LHS to the RHS we get,
$x\large\frac{dv}{dx }= -\frac{(1- v^2) + 2v^2}{2v}$
$x\large\frac{dv}{dx} = -\frac{(1+v^2)}{2v}$
On seperating the variables we get
$2v\large\frac{dv}{(1+v^2) }= -\frac{ dx}{x}$
Step 3:
On integrating both sides we get
$\int \large\frac{2vdv}{(1+v^2)} = - \int\large\frac{ dx}{x }$
$\log(1+v^2) = -\log x +\log C$
$\log\large\frac{(1+v^2)}{x} $$= \log C$
$\large\frac{(1+v^2)}{x} $$= \log C$
Step 4:
Writing the value of $v = \large\frac{y}{x}$ we get
$\large\frac{[1+(y^2/x^2)]}{x}$$ = C$
$x^2+y^2 = Cx$
This is the required solution.
answered Aug 14, 2013 by sreemathi.v

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