Answer to Question #316647 in Differential Equations for Tobias Felix

Question

Bernoulli’s Equation

(y^4 − 2xy)dx + 3x^2 dy = 0 when y(1)=0

Answer 1

$(y^4 − 2xy)dx + 3x^2 dy = 0$ $y(1)=0$

$3x^2\frac{dy}{dx}-2xy=-y^4$

$u=y^{-3}$

$y'=-\frac13u^{-\frac43}u'$

$-3x^2\frac13u^{-\frac43}u'-2xu^{-\frac13}=-u^{-\frac43}$

$x^2u'+2xu=1$

$u'+2\frac {u}{x}=\frac{1}{x^2}$ (1)

Let’s solve the following equation

$u'+2\frac {u}{x}=0$

$\frac {du}{u}=-2\frac{dx}{x}$

$u=C(x)x^{-2}$

To find solution of the equation (1) we should think C is a function of x

$u'=C'x^{-2}-2Cx^{-3}$

Substitution u and u’ into (1) gives

$C'x^{-2}-2Cx^{-3}+2C(x)x^{-3}=\frac{1}{x^2}$

$C'=1$

$C=x+C_1$

$y^{-3}=u=\frac{(x+C_1)}{x^2}$

$y=(\frac{x^2}{x+C_1})^{\frac13}= (\frac{C_2x^2}{C_2x+1})^{\frac13}$

$y(1)=(\frac{C_2}{C_2+1})^{\frac13}=0$

$C_2=0$

$y=0$ .

Answer: $y=0$