Question

The solution of the differential equation $y^{4}dx+2x{y}^{3}dy=\frac{ydx-xdy}{x^3{y}^3}$ is
(where $c$ is integration constant)

• A. $x^3{y}^6+6\ln \frac{y}{x}=c$
• B. $y^3{x}^6+6\ln \frac{y}{x}=c$
• C. $y^3{x}^6+3\ln \frac{y}{x}=c$
• D. $x^3{y}^6+3\ln \frac{y}{x}=c$

Answer 1

The correct option is D $x^3{y}^6+3\ln \frac{y}{x}=c$

The given differential equation can be written as:
$y^{4}dx+2x{y}^{3}dy+\frac{1}{x{y}^3}\frac{(xdy-ydx)}{x^2}=0$
$\Rightarrow x{y}^{7}dx+2x^2{y}^{6}dy+d\left(\frac{y}{x}\right)=0$
Multiplying both sides by $\frac{x}{y},$ we get
$\frac{x}{y}\cdot x{y}^{7}dx+\frac{x}{y}\cdot 2x^2{y}^{6}dy+\frac{x}{y}.d\left(\frac{y}{x}\right)=0$
$\Rightarrow \frac{1}{3}(3x^2{y}^{6}dx+6x^3{y}^{5}dy)+\frac{d\left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)}=0$
$\Rightarrow \frac{1}{3}\left(y^{6}d\right(x^3)+x^{3}d(y^6\left)\right)+\frac{d\left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)}=0$
$\Rightarrow \frac{1}{3}d\left(x^3{y}^6\right)+d\left(\ln \frac{y}{x}\right)=0$
Integrating both sides we get,
$\Rightarrow x^3{y}^6+3\ln \frac{y}{x}=c$