Question

Solution of this differential equation ??Tanydy/dx+tanx=cosycos^2x

Answer 1

check the attachment

Answer 2

ANSWER

=secysecx=sinx +c

STEP-BY-STEP

I am assuming that the differential

equation given by you is: =tany(dydx)+tanx=cos2xcosy

Divide the entire equation by cosy: =tanycosy(dydx)+tanxcosy=cos2x =secytanydydx +tanxsecy=cos2x (1)

Consider:

=1cosy=secy=t=secytanydydx=dtdx

Put,

the substitution in Equation (1),

we get :

= dtdx+tanx×t=cos2x

Equation (1) has now been converted to Equation(2), Which is a linear differential equation of form dtdx+Pt=Q, where P and Q are functions of x

Integrating factor, by definition: =e∫Pdx=e∫tanxdx=secx

Multiplying secx with entire equation we get:d(t×secx)dx=cos2xdx ⇒d(t×secx)=cos2xsecxdx

Solving further:

=t×secx=∫cos2xcosxdx⇒t×secx=∫=cosx

dx⇒t×secx=sinx+c

= secysecx=sinx +c