# Solve the differential equation$\Large \frac{dy}{dx}\normalsize =2xy-y$

$\begin{array}{1 1}(A)\;x=c\;e^{\large y-y^2} \\(B)\;y=c\;e^{\large x-x^2} \\ (C)\;c=xy \\(D)\;x\;e^{\large y-y^2}=c \end{array}$

Toolbox:
• If $\large\frac{dy}{dx}$$=f(x,y), where variables are seperable. • We express it in the form f(x)dx=g(y)dy • Then \int f(x)dx=\int g(y)dy • \int \large\frac{dx}{x}$$=\log |x|+c$
Given $\large\frac{dy}{dx}$$=2xy-y This can be written as \large\frac{dy}{dx}$$=y(2x-1)$
On seperating the variables we get,
$\large\frac{dy}{y}$$=(2x-1)dx Integrating on both sides we get, \int\large\frac{dy}{y}$$=\int (2x-1)dx$
$\log _e y=\bigg( \large\frac{2x^2}{2}-x\bigg)+c$
=>$\log _e y=x^2-x+c$
Converting this to exporential form we get,
$c\;e^{\large x-x^2}=y$
Hence the required equation is
$y=c\;e^{\large x-x^2}$