Question

The solution of $\left(\frac{dy}{dx}\right)(x^2{y}^3+xy)=1$ is:

• A. $\frac{1}{x}=2-y^2+C{e}^{-\frac{y^2}{2}}$
• B. the solution of an equation which is reducible to linear equation.
• C. $\frac{2}{x}=1-y^2+e^{-\frac{y^2}{2}}$
• D. $\frac{1-2x}{x}=-y^2+C{e}^{-\frac{y^2}{2}}$

Answer 1

Answer: (A), (B), (D)

$\frac{1}{x}=2-y^2+C{e}^{-\frac{y^2}{2}}$
the solution of an equation which is reducible to linear equation.
$\frac{1-2x}{x}=-y^2+C{e}^{-\frac{y^2}{2}}$

$\frac{dy}{dx}(x^2{y}^3+xy)=1\Rightarrow \frac{dx}{dy}=x^2{y}^3+xy$
$\Rightarrow \frac{dx}{dy}-xy=y^3{x}^2\Rightarrow -\frac{\frac{dx}{dy}}{x^2}+\frac{y}{x}=-y^3$
Substitute $v=\frac{1}{x}\Rightarrow \frac{dv}{dy}=-\frac{\frac{dx}{dy}}{x{y}^2}$
$\therefore \frac{dv}{dy}+vy=-y^3$
Let $\mu =e^{\int ydy}=e^{\frac{y^2}{2}}$
Multiplying both sides by $\mu$, we get
$e^{\frac{y^2}{2}}\frac{dv}{dy}+e^{\frac{y^2}{2}}vy=-e^{\frac{y^2}{2}}{y}^3$
$\Rightarrow e^{\frac{y^2}{2}}\frac{dv}{dy}+\frac{d}{dx}\left(e^{\frac{y^2}{2}}\right)v=-e^{\frac{y^2}{2}}{y}^3$
Using $g\frac{df}{dy}+f\frac{dg}{dy}=\frac{d}{dy}\left(fg\right)$
$\therefore \frac{d}{dy}\left(e^{\frac{y^2}{2}}v\right)=-e^{\frac{y^2}{2}}{y}^3$
$\Rightarrow \int \frac{d}{dy}\left(e^{\frac{y^2}{2}}v\right)dy=-\int e^{\frac{y^2}{2}}{y}^{3}dy$
$\Rightarrow e^{\frac{y^2}{2}}v=-e^{\frac{y^2}{2}}(y^2-2)+c\Rightarrow v=-y^2+e^{-\frac{y^2}{2}}c+2=\frac{1}{x}$
$\Rightarrow x=-\frac{e^{\frac{y^2}{2}}}{e^{\frac{y^2}{2}}(y^2-2)-c}\Rightarrow \frac{1-2x}{x}=-y^2+c{e}^{-\frac{y^2}{2}}$