Answer
See answers below
Work Step by Step
a) The Riccati differential equation:
$$y'+7x^{-1}y-3y^2=3x^{-2}$$
We have $y=x^{-1}+w$
Derivative:
$$y'=-x^{-2}+w'(x)$$
Substituting:
$$-x^{-2}+w'+7x^{-1}(x^{-1}+w)-3(x^{-1}+w)^2=3x^{-2}$$
$$-x^{-2}+w'+7x^{-2}+7x^{-1}w-3x^{-2}-6x^{-1}w-3w^2=3x^{-2}$$
It is equivalent with the Bernoulli equation:
$$w'+x^{-1}w=3w^2$$
b) Dividing the Bernoulli equation by $w^2$
$$w^{-2}w'+x^{-1}w^{-1}=3 (1)$$
With $v=w^{-1} \rightarrow v'=-w^{-2}w'$
The equation (1) becomes:
$$v'-x^{-1}v=-3$$
The integrating factor is:
$$I=e^{-\int x^{-1}}=e^{-\ln x}=x^{-1}$$
Then,
$$\frac{d}{dx}(x^{-1}v)=-3x^{-1}$$
Integrating both sides:
$$x^{-1}v=-3\ln x+C$$
where C is a constant of integration
$$\rightarrow v=x(C-3\ln x)$$
$$\rightarrow w=\frac{1}{x(C-3\ln x)}$$
The final solution is:
$$y=\frac{1}{x}(1+\frac{1}{C-3\ln x})$$
