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Solve the following $\large\frac{dy}{dx}$=$\large\frac {y(x-2y)}{x(x-3y)}$

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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)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
  • To solve we put $y=vx$ and proceed.
$\large\frac{dy}{dx}=\frac{y(x-2y)}{x(x-3y)}$
This is of the form $ \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
We substitute $y=vx=>\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes $v+x \large\frac{dv}{dx} = \frac{vx.x(1-2v)}{x.x(1-3v)}$
$v+x\large\frac{dv}{dx}=\frac{v(1-2v)}{1-3v}$
$x \large\frac{dv}{dx}=\large \frac{v(1-2v)-v(1-3v)}{1-3v}$
$\qquad= \large\frac{v-2v^2-v+3v^2}{1-3v}$
$\qquad=\large\frac{v^2}{1-3v}$
$\therefore \large \frac{1-3v}{v^2}$$dv=\large\frac{dx}{x}$
Step 3:
The variables are separated
$\int \bigg(\large\frac{1}{v^2}-\frac{3}{v}\bigg)$$dv=\int \large\frac{dx}{x}$$-c$
$-\large\frac{1}{v}$$- 3 \log v=\log x -c_1$
Substituting $xv=\large\frac{y}{x}$
$c_1-3 \log v -\log x=\large\frac{1}{v}$
$\log c-\log v^3 -\log x =\large\frac{1}{v}$
$\log \large\frac{c}{v^3x}=\large\frac{1}{v}$
Substituting $v=\large\frac{y}{x}$
$\log \large\frac{cx^3}{y^3x}=\frac{x}{y}$
$\large\frac{cx^2}{y^3}=e^{\Large\frac{x}{y}}$
or $y^3=cx^2e^{\Large\frac{-x}{y}}$
answered Sep 5, 2013 by meena.p
 

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