Browse Questions

# Solve:$y\;dx-x\;dy=x^2y\;dx$

$\begin{array}{1 1} (A)\;y=c\;e^{\Large\frac{-x^2}{2}+x} \\ (B)\;x=c\;e^{\Large\frac{-x^2}{2}+y} \\(C)\;xc\;e^{\Large\frac{-x^2}{2}+x}=0 \\ (D)\;yc\;e^{\Large\frac{-x^2}{2}+x}=0\end{array}$

Toolbox:
• A linear differential equation of the form.
• $\large\frac{dy}{dx}$$=f(x), can be solved by seperating the variable and then integrating it. • \int \large\frac{dx}{x}$$=\log |x|$
Given $ydx-xdy=x^2ydx$
This can be written as
$-xdy=x^2ydx-ydx$
=>$-xdy=ydx(x^2-1)$
Now seperating the variables we get,
$\large\frac{dy}{y}$$=-\bigg(\large\frac{x^2-x}{x}\bigg)$$dx$
=>$\large\frac{dy}{y}$$=(-x+1)dx Integrating on both sides we get \int \large\frac{dy}{y}$$=\int -xdx+\int dx$
=>$\log e^y=-\large\frac{x^2}{2}$$+x+c$
$e^{\Large\frac{-x^2}{2}+x}+c=y$
$y=c\;e^{\Large\frac{-x^2}{2}+x}$

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