Question

The solution of the differential equation $\frac{dy}{dx}=\frac{3x^{2}y^{4}+2xy}{x^{2}-2x^{3}\,y^{3}}$ is

• A. $\frac{y^{2}}{x}-x^{3}y^{2}=c$
• B. $\frac{x^{2}}{y^{2}}+x^{3}y^{3}=c$
• C. $\frac{x^{2}}{y}+x^{3}y^{2}=c$
• D. $\frac{x^{2}}{3y}-2 x^{3}y^{2}=c$

Answer 1

Answer: (C)

Rewrite the differential equation as
$(2xy\,dx-x^{2}\,dy)+y^{2}(3x^{2}\,y^{2}\,dx+2x^{3}y\,dy)=0$
Dividing by $y^{2}$, we get
$\frac{y\,2x\,dx-x^{2}\,dy}{y^{2}} +y^{2}\,3x^{2}\,dx+x^{3}\,2y\,dy=0$
or $d=\left(\frac{x^{2}}{y}\right)+d \left(x^{3} y^{2}\right)=0$
Integrating, we get the solution $\frac{x^{2}}{y}+x^{3}y^{2}=c$