Answer to Question #106401 in Differential Equations for khushi

Question

Find the solution of the Riccati equation
dy/dx=(2cos^2(x)-sin^2(x)+y^2)/2cos(x); y1(x)=sinx

Answer 1

Find the solution of the Riccati equation

dy/dx=(2cos^2(x)-sin^2(x)+y^2)/2cos(x); y1(x)=sinx

Solution:

$y'=(2cos^2x-sin^2x+y^2)/2cos(x)\newline y' =(0.5sec(x))y^2+(2cos^2(x)-sin^2(x))/2cos(x)$

$y1=sinx$

riccati eqyation general form

$y'(x) = A(x)y^2 + B(x)y + C(x)...(1)\newline A(x) =0.5sec(x)\newline B(x)=0\newline C(x)=(2cos^2(x)-sin^2(x))/2cos(x)$

now substituting

$y(x)=v(x)+y_1(x)$

$y'(x)=v'(x)+y_1'(x)\newline$

in equation (1)

$y_1 \hspace{0.1cm}is \hspace{0.1cm}solution\hspace{0.1cm} to \hspace{0.1cm}above \hspace{0.1cm}differential \hspace{0.1cm}equation\hspace{0.1cm} \newline y _ 1' = A(x)y ^2_ 1 + B(x)y_1 + C(x).$

$v'+y_1'== A(x)[v + y1]^2 + B(x)[v + y1 ] + C(x)\newline ⇒ v'+ [A(x)y^2_ 1 + B(x)y_1 + C(x) ] = A(x)v^2 + 2A(x)y_1v + A(x)y^ 2_ 1 +B(x)v + B(x)y_1 + C(x)\newline ⇒ v' = A(x)v^ 2 + 2A(x)y_1v + B(x)v\newline \newline ⇒ v' + [−2A(x)y_1(x) − B(x)]v = A(x) v^2...(2) \newline \newline equation\hspace{0.1cm} (2)\hspace{.1cm} is \hspace{.1cm}bernouli \hspace{.1cm} equation \newline v'+p(x)v=q(x)v^2\newline p(x) = [−2A(x)y_1(x) − B(x)]=-2tan(x) \newline q(x) = A(x)=0.5sec(x) \newline$

solving this bernouli equation

$v'-2tan(x)v=0.5sec(x)v^2\newline let\hspace{0.1cm}z=1/v\implies z'=-1/v^2\newline z'+2ztan(x)=-0.5sec(x)\newline this \hspace{0.1cm} is \hspace{0.1cm} linear \hspace{0.1cm} differential \hspace{0.1cm} equation\newline IF=exp(\intop2tan(x)dx)=sec^2x \newline \therefore [sec^2(x).z]'=-0.5sec(x)\newline z=(tan(x)sec(x)+ln|tan(x)+sec(x)|)/(-4sec^2(x))\newline \because z=1/v\newline \newline v=-4sec^2(x)/(tan(x)sec(x)+ln|tan(x)+sec(x)|)\newline y=v+y_1\newline y=-4sec^2(x)/(tan(x)sec(x)+ln|tan(x)+sec(x)|)+sinx$

Answer:

$y=-4sec^2(x)/(tan(x)sec(x)+ln|tan(x)+sec(x)|)+sinx$