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

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