Question

The solution of the differential equation $\frac{dy}{dx}=\frac{1}{x+y^2}$ is

• A. $y=-x^2-2x-2+c{e}^x$
• B. $y=x^2+2x+2-c{e}^x$
• C. $x=-y^2-2y+2-c{e}^y$
• D. $x=-y^2-2y-2+c{e}^y$
• E. $x=y^2+2y+2-cey$

Answer 1

Answer: (D)

Given differential equation is $\frac{dy}{dx}=\frac{1}{x+y^2}$
$\Rightarrow$ $\frac{dx}{dy}-x=y^2$ Here, $P=-1,Q=y^2$ If
$=e^{\int -1dy}=e^{-y}$
$\therefore$ Solution is $x{e}^{-y}=\int e^{-y}{y}^{2}dy$
$=-e^{-y}{y}^2+\int 2e^{-y}ydy$
$=-e^{-y}{y}^2+2[-e^{-y}y+\int e^{-y}dy]+c$
$=-e^{-y}{y}^2+2[-e^{-y}y-e^{-y}]+c$
$\Rightarrow$ $x{e}^{-y}=e^{-y}(-y^2-2y-2)+c$
$\Rightarrow$ $x=-y^2-2y-2+c{e}^y$