найти решение дифуравнения y``+4y= 1/sin2x y(п/4)=2, у`(п/4)=п/4

y'' + 4y = 0

(a^2) + 4 = 0
a^2 = - 4
a = -2i; a = 2i

y1(x) = sin2x
y2(x) = cos2x

y1(x) и y2(x) образуют фундаментальную систему решений

(y1)' = 2cos2x
(y2)' = - 2sin2x

y'' + 4y = 4/(sin2x)

y(x) = C(x)sin2x + D(x)cos2x

C(x) и D(x) находятся из системы
C'y1 + D'y2 = 0; C'y1' + D'y2' = 4/(sin2x)
C'(sin2x) + D'(cos2x) = 0; 2C'(cos2x) - 2D'(sin2x) = 4/(sin2x)

C' = - D'(cos2x)/(sin2x)

2C'(cos2x) - 2D'(sin2x) = 4/(sin2x)
C'(cos2x) - D'(sin2x) = 2/(sin2x)
- D'((cos2x)^2)/(sin2x) - D'(sin2x) = 2/(sin2x)
- D'((cos2x)^2) - D'((sin2x)^2) = 2
D'((cos2x)^2 + (sin2x)^2) = - 2
D' = - 2
D(x) = - 2x + A

C' = - D'(cos2x)/(sin2x) = 2(cos2x)/(sin2x)
C(x) = int 2(cos2x)dx/(sin2x) = int d(sin2x)/(sin2x) =
= ln|sin2x| + B

y(x) = C(x)sin2x + D(x)cos2x

y(x) = (ln|sin2x| + B)(sin2x) + (-2x + A)(cos2x)

y(П/4) = 2
2 = (ln1 + B)*1 + (-П/2 + A)*0
2 = 0 + B
2 = B

y(x) = (ln|sin2x| + B)(sin2x) + (-2x + A)(cos2x)
y(x) = (ln|sin2x| + 2)(sin2x) + (-2x + A)(cos2x)

y'(x) = 2cos2x + 2(ln|sin2x| + 2)(cos2x) - 2cos2x -
- 2(-2x + A)(sin2x) =
= 2(ln|sin2x| + 2)(cos2x) + (4x - 2A)(sin2x)

y'(П/4) = П
П = 2(ln1 + 2)*0 + (П - 2A)*1
П = П - 2A
A = 0

y(x) = (ln|sin2x| + 2)(sin2x) + (-2x + A)(cos2x)
y(x) = (ln|sin2x| + 2)(sin2x) + -2x(cos2x)