Respuesta :
The general solution of equation y''+4y=sec(2z) is
y = c₁cos2z + c₂sin2z + [tex]\frac{cos2z}{4}[/tex] ln(cos2z) + [tex]\frac{z sin2z}{2}[/tex].
Given differential equation is,
y'' + 4y = sec(2z)
Characteristic equation of this equation is
(D²+4)=0
⇒ D² = -4
D = ±2i
Therefore the roots are imaginary.
D₁=2i ⇒ y₁ = e⁰cos2z = cos2z
D₂=-2I⇒ y₂ = e⁰sin2z = sin2z
yn= c₁cos2z + c₂sin2z
Noe to solve yp, first we need to solve Wronskian
y₁= cos2z , y₂=sin2z
y₁'= -2sin2z , y₂'=2cos2z
⇒ W = [tex]\left[\begin{array}{ccc}y1&y2\\y1'&y2'\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}cos2z&sin2z\\-2sin2z&2cos2z\\\end{array}\right][/tex]
= (cos2z)(2cos2z)-(-2sin2z)(sin2z)
= 2cos²2z + 2sin²2z
=2 (cos²2z + sin²2z)
= 2 ≠ 0
yp = - y₁[tex]\int\ {\frac{y2 g(z)}{W} } \, dz[/tex] + y₂[tex]\int\ {\frac{y1 g(z)}{W} } \, dz[/tex]
= -(cos2z)[tex]\int\ {\frac{(sin2z)(sec2z)}{2} } \, dz[/tex] + (sin2z)[tex]\int\ {\frac{(cos2z)(sec2z)}{2} } \, dz[/tex]
= - (cos2z)[tex]\int\ {\frac{sin2z/cos2z}{2} } \, dz[/tex] + (sin2z)[tex]\int\ {\frac{cos2z/cos2z}{2} } \, dz[/tex]
= - [tex]\frac{cos2z}{2}[/tex][tex]\int\ {tan2z} \, dz[/tex] + [tex]\frac{sin2z}{2}[/tex][tex]\int\ \,dz[/tex]
= - [tex]\frac{cos2z}{2}[/tex](-[tex]\frac{1}{2}[/tex] ln (cos2z)) + [tex]\frac{sin2z}{2}[/tex] (z)
= [tex]\frac{cos2z}{4}[/tex]ln(cos2z) + [tex]\frac{zsin2z}{2}[/tex]
Now the general equation will become
y = yn + yp
y = c₁cos2z + c₂sin2z +[tex]\frac{cos2z}{4}[/tex]ln(cos2z) + [tex]\frac{zsin2z}{2}[/tex] .
To know more about Wronsian here
https://brainly.com/question/16930425
#SPJ4
