Question

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

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

Answer 1

The correct option is B $x^3{y}^2+\frac{x^2}{y}=c$


$x^{2}dy-2x^3{y}^{3}dy=3x^2{y}^{4}dx+2xydx\Rightarrow x^{2}dy-2xydx=3x^2{y}^{4}dx+2x^3{y}^{3}dy\Rightarrow \frac{2xydx-x^{2}dy}{y^2}+3x^2{y}^{2}dx+2x^{3}ydy=0\Rightarrow d\left(\frac{x^2}{y}\right)+d\left(x^3{y}^2\right)=0\Rightarrow \frac{x^2}{y}+x^3{y}^2=C$