Given, cosec2xdy+2dx=(1+ycos2x)cosec2xdx
⇒sin2xdy+2dx=(sin2x1+ycos2x)dx
⇒dxdy+2sin2x=1+ycos2x
⇒dxdy+2sin2x−1=ycos2x
Using, cos2x=1−2sin2x, we get
dxdy+(−cos2x)=ycos2x
⇒dxdy+(−cos2x)y=cos2x
This is a linear differential equation of the type dxdy+Py=Q, where P and Q are the functions of x or constants.
Thus, P=−cos2x and Q=cos2x
Now, we have integrating factor I.F.=e∫Pdx=e∫−cos2xdx=e−(2sin2x)
And, the solution of the given differential equation is
y(I.F.)=∫Q(I.F.)dx+c
⇒y⋅e−(2sin2x)=∫(cos2x)⋅e−(2sin2x)dx+c
Put 2sin2x=t,⇒22cos2xdx=dt
⇒y⋅e−(2sin2x)=∫e−tdt+c
⇒y⋅e−(2sin2x)=−e−tdt+c
⇒y⋅e−(2sin2x)=−e−(2sin2x)+c
Given, y(4π)=0
⇒0⋅e−2sin(2π)=−e−2sin(2π)+c
⇒c=−e−(21)
⇒y⋅e−(2sin2x)=−e−(2sin2x)+e−21
Now, at x=0
⇒y(0)⋅e−(2sin0)=−e−(2sin0)+e−21
⇒y(0)⋅e0=−e0+e−21
⇒y(0)=−1+e−21
⇒y(0)+1=e−21
⇒(y(0)+1)2=e−1.
