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The solution of the differential equation $y dx + (x + xy) dy = 0$ is

$\begin{array}{1 1} (A)\;ce^{y}=xy \\ (B)\;ce^{-y}=xy \\ (C)\;ce^{-y}=x+y \\ (D)\;ce^{-y}=x-y\end{array}$

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x),$ can be solved by seperating the variables and then integrating it
  • $\int \large\frac{1}{x}$$=\log |x|+c$
$ydx+(x+xy)dy=0$ is
This can be written as
(ie) $x(1+y)dy=-ydx$
Seperating the variables we get,
$\large\frac{(1+y)dy}{y}=- \frac{dx}{x}$
Now seperating the terms and integrating,
$\int \large\frac{1}{y} $$dy+\int dy=-\int \large\frac{dx}{x}$
$=>\log y+y=-\log x+c$
(ie) $\log y+y=-\log x+c$
$=>\log x+\log y=-y+c$
(or) $\log xy=-y+c$
$ce^{-y}=xy$ is the required solution .
Hence the solution of the differential equation
$ydx+(x+xy)dy=0$ is $ce^{-y}=xy$
answered May 22, 2013 by meena.p