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

1 Answer

  • A function is said to be homogenous function in degree n if $F(kx,ky) = k^nF(x,y)$ for any non zero constant $k.$
  • To solve these type of homogenous functions we make the substitution $y = vx$ hence $\large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
Given: $(x^2 + xy)dy = (x^2 + y^2)dx$
$\large\frac{dy}{dx} =\frac{ (x^2 + y^2)}{(x^2 + xy)}$
$F(x,y) =\large\frac{ (x^2 + y^2)}{(x^2+xy)}$
$F(kx,ky) = \large\frac{k[x^2+y^2]}{[x^2 + xy]}$$ = k^0 $
Hence this is a homogenous function of degree 0.
Step 2:
Using the information in the tool box let us substitute for $y = vx$ and $\large\frac{dy}{dx}$$ = v + x\large\frac{dv}{dx}$
$v + x\large\frac{dv}{dx} =\frac{ [x^2 +x^2 v^2]}{x^2 + xvx}$
Taking the common factor $x^2$ and cancelling we get
$v + x\large\frac{dv}{dx} =\frac{ [1+v^2]}{1+v}$
bringing $v$ from the LHS to the RHS and simplifying we get,
$x\large\frac{dv}{dx} =\frac{ 1-v}{1+v}$
on seperating the variables we get,
$\large\frac{[1+v]}{[1-v]} =\frac{ dx}{x}$
$\large\frac{2}{[1-v] - 1} =\frac{ dx}{x}$
Step 3:
On integrating both sides we get,
$-2\log[1-v] -v = \log x + \log c$
$v = -2\log[1-v] - \log x +\log C$
$v=\large\frac{\log c}{x(1-v)^2}$
substituting for $v$ we get
$\large\frac{y}{x} =\frac{ \log C}{x[1-(y/x)^2]}$
$\large\frac{cx}{(x-y)^2 }=$$e^{\Large (y/x)}$
on rearranging we get
$(x-y)^2= Cxe^{\Large(-y/x)}$
This is the required solution.
answered Aug 14, 2013 by sreemathi.v

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