Question

The integrating factor for the Bernouli differential equation 2xyy' = y^2 - 2x^3 given initial condition y(1)=2

Answer 1

Rewrite in the form of a first order ODE

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

Substitute $z=y^{1-(-1)}=y^2$

$$z'=2yy'$$

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

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

Integrating factor

$$IF=\mu(x)=e^{\int(-1/x)dx}=\dfrac{1}{x}$$

$$\dfrac{1}{x}z'-\dfrac{1}{x}(\dfrac{1}{x})z=-2x^2(\dfrac{1}{x})$$

$$d(\dfrac{z}{x})=-2xdx$$

Integrate

$$\int d(\dfrac{z}{x})=-\int 2xdx$$

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

$$z=-x^3+Cx$$

$$y^2=-x^3+Cx$$

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

Given initial condition $y(1)=2$

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

$$C=9$$

$$y=\sqrt{-x^3+9x}$$