Question

Solve the differential equation: $\frac{dy}{dx}=\frac{y^3+3x^{2}y}{x^3+3x{y}^2}$

• A. $xy=k^2(x^2-y^2{)}^2$
• B. $xy=k^2(x^2+y^2{)}^2$
• C. $xy=-k^2(x^2+y^2{)}^2$
• D. $xy=-k^2(x^2-y^2{)}^2$

Answer 1

Answer: (A), (D)

$xy=k^2(x^2-y^2{)}^2$
$xy=-k^2(x^2-y^2{)}^2$

$\frac{dy}{dx}=\frac{y^3+3x^{2}y}{x^3+3x{y}^2}$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\therefore x\frac{dv}{dx}+v=\frac{(v^2+3)v}{3v^2+1}$

$\Rightarrow \frac{dv}{dx}=\frac{-2(v^2-1)v}{x(3v^2+1)}\Rightarrow \frac{3v^2+1}{v(v^2-1)}\frac{dv}{dx}=-\frac{2}{x}$

Integrating both sides w.r.t $x$ we get

$\int \frac{3v^2+1}{v(v^2-1)}\frac{dv}{dx}dx=-\int \frac{2}{x}dx$

$\frac{3v^2+1}{v(v^2-1)}=\frac{3v^2}{v(v^2-1)}+\frac{1}{v(v^2-1)}=\frac{3v}{v^2-1}+\frac{1}{v(v^2-1)}=\frac{3}{2(v-1)}+\frac{3}{2(v+1)}-\frac{1}{v}+\frac{1}{2(v-1)}+\frac{1}{2(v+1)}$(using partial fractions)

After integrating we will get

$\Rightarrow 2\log (1-v^2)-\log v=-2\log x+c\Rightarrow xy=\pm k{(x^2-y^2)}^2$