Answer to Question #217658 in Differential Equations for okaj

Question

y'=(2cos^2(x)-sin^2(x)+y^2)/2cos(x), y(0)=-1, y(1)=sin(x)

Answer 1

Solution;

The equation is Ricatti of the form;

y'=A(x)y²+B(x)y+C(x)

From our equation,

A(x)=$\frac{1}{2cos(x)}$

B(x)=0

C(x)=$\frac{2cos^2(x)-sin^2(x)}{2cos(x)}$

Given y₁(x)=sin (x)

To find the general solution,

Let y=y₁+$\frac 1 v$

Where v is another function of x.

y=sin(x)+$\frac 1v$

y'=cos(x)-$\frac {1}{v^2}$$\frac{dv}{dx}$

We can now equate the equations,

$\frac{dy}{dx}$ =cos(x)-$\frac{1}{v^2}\frac{dv}{dx}$ =$\frac{2cos^2(x)-sin^2(x)+(sin(x)+\frac 1v)^2}{2cos(x)}$

By simplication;

cos(x)-$\frac{1}{v^2}\frac{dv}{dx}$ =$\frac{2cos^2(x)}{2cos(x)}+\frac{\frac{2sin(x)}{v}+\frac{1}{v^2}}{2cos(x)}$

Further simplification;

$-\frac{1}{v^2}\frac{dv}{dx}=\frac{\frac{2sin(x)}{v}+\frac{1}{v^2}}{2cos(x)}$

Multiply -v²

$\frac{dv}{dx}$ =$\frac{-(2vsin(x)+1)}{2cos(x)}=-vtan(x)$ -0.5sec(x)

Which be rewritten as;

$\frac{dv}{dx}$ +vtan (x)=-0.5sec(x)

Introduce the integrating factor,I.F

I.F=$e^{\int{tan(x)dx}}$ =sec(x)

Multiply all through with I.F;

sec(x)$\frac{dv}{dx}$ +vsec(x)tan(x)=-0.5sec²(x)

D_x(vsec(x)=-0.5sec²(x)dx

Integrate both sides;

vsec(x)=-0.5tan(x)+C

Multiply both sides with cos(x);

V=-0.5sin(x)+Ccos(x)

y=sin(x)+$\frac 1v$

y=sin(x)+$\frac{1}{Ccos(x)-0.5sin(x)}$

Using the given intial condition,find the value of C;

y(0)=-1

-1=sin(0)+$\frac{1}{Ccos(0)-0.5sin(0)}$

-1=$\frac1C$

C=-1

Therefore,the general solution of the equation is ;

y=sin(x)-$\frac{1}{cos(x)+0.5sin(x)}$

This can be written as;

y=sin(x)-$\frac{2}{2cos(x)+sin(x)}$