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Solve the following $(x^{2}+y^{2})dx+3xy dy$=$0$

1 Answer

  • 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)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)}$ Where $f_1$ and $f_2$ are homogeneous functions in x and y.
  • To solve we put $y=vx$ and proceed.
Step 1:
$\large\frac{dy}{dx} =\frac{-(x^2+y^2)}{3xy}$ is homogeneous in $x,y$
$y=vx$=> $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes
$v+ x \large\frac{dv}{dx}=\large\frac{-(x^2+v^2x^2)}{3x.vx}$
$v+ x \large\frac{dv}{dx}=\large\frac{-(1+v^2)}{3.v}$
$ x \large\frac{dv}{dx}=-\bigg(\large\frac{1+v^2}{3v}\bigg)$$-v$
$\qquad= \large\frac{-1-v^2-3v^2}{3v}$
$\qquad= \large\frac{-(1+4v^2)}{3v}$
Step 3:
The variables are separated
$\int \large\frac{3v}{1+4v^2}$$dv=-\int \large\frac{dx}{x}$$+\log c_1$
$\large\frac{3}{8} \int \large\frac{8v}{1+4v^2}$$dv = - \int\large\frac{dx}{x}$$+\log c_1$
$\large\frac{3}{8}$$ \log (1+4v^2)+\log x=\log c_1$
Substitute $v=\large\frac{y}{x}$
$\bigg(1+4 \large\frac{y^2}{x^2}\bigg)^{3/8}$$x=c_1$
$(x^2+4y^2)^{3/8}. x^{1/4}=c_1$
$(x^2+4y^2)^3.x^2=c$ is the G.S


answered Sep 5, 2013 by meena.p