Question #105501
b) Find the solution of the Riccati equation
dy/dx=(2cos^2x-sin^2x+y^2)/2cosx ;y(x)=sinx
1
Expert's answer
2020-03-24T11:41:39-0400

The general for Ricatti's equation:

y=P(x)+Q(x)y+R(x)y2y'=P(x)+Q(x)y+R(x)y^2

Here

P(x)=2cos2xsin2x2cosx;Q(x)=0;R(x)=12cosxP(x)=\frac{2\cos^2x-\sin^2x }{2\cos x}; Q(x)=0;R(x)=\frac{1}{2\cos x}

Since y1=sinxy_1=\sin x is the particular solution of the equation, we can build the general solution of the equation in form

y=y1+1w(x)y=y_1+\frac{1}{w(x)}

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

w=(Q(x)+2R(x)y1)wR(x)w'=-(Q(x)+2R(x)y_1)w-R(x)

or

w=sinxcosxw12cosxw'=-\frac{\sin x}{\cos x}w-\frac{1}{2\cos x}

w+wtanx=12cosxw'+w\tan x=-\frac{1}{2\cos x}

This is a first order linear equation, which solution is

w=c1cosxsinx2w=c_1\cos x-\frac{\sin x}{2}

Therefore, the general solution of the equation is

y=sinx+2Ccosxsinx,C=2c1y=\sin x+\frac{2}{C\cos x-sinx}, C=2c_1


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