Answer to Question #215112 in Differential Equations for shai

Question

Solve the equation (Bernoulli Differential Equation)

(2y^3 - x^3) dx + 3x^2y dx = 0

Answer 1

$(2y^3 - x^3) dx + 3x^2y dy = 0$

$\frac{dx}{dy}=\frac{-3x^2y}{2y^3-x^3}=\frac{-3}{\frac{2y^2}{x^2}-\frac{x}{y}}$

$\frac{y}{x}=v$

y=vx

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

$x\frac{dv}{dx}=\frac{-3v}{2v^3-1}-v=\frac{-2v-2v^4}{2v^3-1}$

$\int\frac{2v^3-1}{-2v-2v^4}dv=\int\frac{dx}{x}$

$-2v-2v^4=-v(2v^3-1)+3)$

$-2v-2v^4=t$

$(-2-8v^3)dv=dt$

$-4(2v^3-1+3)dv=dt$

$-4\int\frac{dt}{t}-4\int3dv=\int\frac{dx}{x}$

-4log(t)-12v=log(x)+log(c)

$-4log(\frac{t}{xc})-12v=-4log(\frac{-2v(1+v^3)}{xc})-12v=-4log(\frac{-2\frac{y}{x}(1+(\frac{y}{x})^3}{xc})-12\frac{y}{x}=0$

$\frac{-2y(1+(\frac{y}{x})^3)}{x^2c}=e^{-3\frac{y}{x}}$