Answer to Question #248059 in Mechanical Engineering for ramyr

Question

(4x^3-3y^2)dx + 6xydy=0,f(-1)=0

Answer 1

$6xyy'-3y^2=-4x^3$

$y'-\dfrac{1}{2x}y=-\dfrac{2}{3}x^2y^{-1}$

$v=y^{1-n}=y^{1-(-1)}=y^2$

$v'=2yy'$

$\dfrac{1}{2}v'-\dfrac{1}{2x}v=-\dfrac{2}{3}x^2$

$v'-\dfrac{1}{x}v=-\dfrac{4}{3}x^2$

The integrating factor

$\mu(x)=\dfrac{1}{x}$

$\dfrac{1}{x}v'-\dfrac{1}{x^2}v=-\dfrac{4}{3}x$

$d(\dfrac{v}{x})=-\dfrac{4}{3}xdx$

Integrate

$\int d(\dfrac{v}{x})=-\int \dfrac{4}{3}xdx$

$\dfrac{v}{x}=-\dfrac{2}{3}x^2+C$

$v=-\dfrac{2}{3}x^3+Cx$

$y^2=-\dfrac{2}{3}x^3+Cx$

$y=\pm\sqrt{-\dfrac{2}{3}x^3+Cx}$

Given $y(-1)=0$

$0=\pm\sqrt{-\dfrac{2}{3}(-1)^3+C(-1)}$

$C=\dfrac{2}{3}$

Answer:

$y(x)=-\sqrt{-\dfrac{2}{3}x^3+\dfrac{2}{3}x}$

or

$y(x)=\sqrt{-\dfrac{2}{3}x^3+\dfrac{2}{3}x}$