Question

b) Find the solution of the Riccati equation

dy/dx=(2cos^2x-sin^2x+y^2)/2cosx ;y(x)=sinx

Answer 1

The general for Ricatti's equation:

$y'=P(x)+Q(x)y+R(x)y^2$

Here

$P(x)=\frac{2\cos^2x-\sin^2x }{2\cos x}; Q(x)=0;R(x)=\frac{1}{2\cos x}$

Since $y_1=\sin x$ is the particular solution of the equation, we can build the general solution of the equation in form

$y=y_1+\frac{1}{w(x)}$

Where $w(x)$ is the solution to the first-order linear equation

$w'=-(Q(x)+2R(x)y_1)w-R(x)$

or

$w'=-\frac{\sin x}{\cos x}w-\frac{1}{2\cos x}$

$w'+w\tan x=-\frac{1}{2\cos x}$

This is a first order linear equation, which solution is

$w=c_1\cos x-\frac{\sin x}{2}$

Therefore, the general solution of the equation is

$y=\sin x+\frac{2}{C\cos x-sinx}, C=2c_1$