Question

$\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}x\frac{\mathrm{dy}}{\mathrm{dx}}+y={\mathrm{xy}}^3?$

• A. $y^2=\frac{1}{2x+{\mathrm{cx}}^2}$
• B. $y^{}=\frac{1}{2x+{\mathrm{cx}}^2}$
• C. $y^2=2x+{\mathrm{cx}}^2$
• D. $y^3=\frac{1}{2x+{\mathrm{cx}}^2}$

Answer 1

The correct option is A $y^2=\frac{1}{2x+{\mathrm{cx}}^2}$

$x\frac{\mathrm{dy}}{\mathrm{dx}}+y={\mathrm{xy}}^3\mathrm{Dividing}\mathrm{by}{\mathrm{xy}}^3\frac{1}{y^3}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{{\mathrm{xy}}^2}=1\mathrm{Take}\frac{1}{y^2}=u\mathrm{Differentiating}\mathrm{wrt}x\frac{-2}{y^3}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}\frac{1}{y^3}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{du}}{2\mathrm{dx}}-\frac{\mathrm{du}}{2\mathrm{dx}}+\frac{u}{x}=1\frac{\mathrm{du}}{\mathrm{dx}}-\frac{2u}{x}=-2\mathrm{Integrating}\mathrm{factor}=e^{{\int }^{}\frac{-2}{x}\mathrm{dx}}=e^{-2\mathrm{lnx}}=e^{\ln \left(x^{-2}\right)}=x^{-2}u=x^2\int \frac{-2}{x^2}\mathrm{dx}+x^{2}cu=x^2\frac{2}{x}+{\mathrm{cx}}^{2}u=2x+{\mathrm{cx}}^2\mathrm{Substituting}u\frac{1}{y^2}=2x+{\mathrm{cx}}^2{y}^2=\frac{1}{2x+{\mathrm{cx}}^2}$