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Solve the following $(x^{2}+y^{2})\;dy$ =$ xy \;dx$

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  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate then, ie
  • $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • Can be written as $ \large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
  • A D.E of first order and first degree is said to be homogeneous if it can be put in the form $\large\frac{dy}{dx}=f\bigg(\large\frac{y}{x}\bigg)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
  • To solve we put $y=vx$ and proceed.
Step 1:
$(x^2+y^2)dy=xy \;dx$
$\large\frac{dy}{dx}=\large\frac{xy}{x^2+y^2}$
This is of the form $ \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
Let $y=vx=>\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes $v+ x \large\frac{dv}{dx}=\frac{x-vx}{x^2+v^2x^2}$
$\qquad=\large\frac{v}{1+v^2}$
$x \large\frac{dv}{dx}=\large\frac{v}{1+v^2}$$-v$
$\qquad=\large\frac{ v-v(1+v^2)}{1+v^2}$
$\qquad= \large\frac{-v^3}{1+v^2}$
$\large\frac{1+v^2}{v^3}$$dv=\large\frac{-dx}{x}$
Step 3:
The variables are separated
$\int \large\frac{1+v^2}{v^3}$$dv=-\int \large\frac{dx}{x}$$+\log c$
$\int \bigg(\large\frac{1}{\sqrt 3}+\frac{1}{v}\bigg)$$dv=-\int \large\frac{dx}{x}$$+\log c$
$\large\frac{-1}{2} \frac{1}{\sqrt 2}$$+\log v=-\log x +\log c$
$\log x+\log v-\log c=\large\frac{1}{2v^2}$
$\log \large\frac{xv}{c}=\large\frac{1}{2v^2}$
=>$\large\frac{xv}{c}=e^{\Large\frac{-1}{2v^2}}$
Substituting $v=\large\frac{y}{x}$ the GS becomes
$y=ce^{\Large\frac{-x^2}{2y^2}}$
answered Sep 4, 2013 by meena.p
 

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