# The solution of the differential equation $\large\frac{dy}{dx}$$=e^{x-y}+x^2e^{-y} is :$(A)\;y=e^{x-y}-x^2e^{-y}+c \quad (B)\;e^y-e^x=\frac{x^3}{3}+c \quad(C)\;e^x+e^y=\frac{x^3}{3}+c \quad (D)\;e^x-e^y=\frac{x^3}{3}+c$ ## 1 Answer Toolbox: • If a linear differential equation is of the form \large\frac{dy}{dx}$$=f(x),$ then it can be solved by seperating the variables.
• $\int e^xdx=e^x+c$
Given $\large\frac{dy}{dx}$$=e^{x-y}+x^2e^{-y} This can be written as \large\frac{dy}{dx}=\Large\frac{e^x}{e^y}+\frac{x^2}{e^y} \large\frac{dy}{dx}=\Large\frac{e^x+x^2}{e^y} On seperating the variables we get e^y.dy=(x^2+e^x)dx On integrating we get, \int e^y dy=\int x^2 dx+\int e^x dx e^y=\large\frac{x^3}{3}$$+e^x+c$
$e^y-e^x=\large\frac{x^3}{3}+c$
Hence the correct option is $B$