inver persamaan diferensial biasa

10.  y”+ 4y = sin² 2x

Jawab :

Ø  Solusi PDH

PDH     =  y”+ 4y = 0

PK        = λ ² + 4 = 0

ADK     = λ  = ± 2i

Ø  Solusi Yh

Yh        = C₁ cos 2x + C₂ sin 2x

Ø  Solusi Yp

Yp        = U₁Y₁ + U₂Y₂

                        = U₁cos2x + U₂sin2x pmm2

Fungsi U₁,U₂ diperoleh dari :

cos2xsin2x-2sin2x2cos2xu1'u2' = 0sin2 2x

W = cos2xsin2x-2sin2x2cos2x=2

W₁ = 0sin2xsin2 2x2cos2x=-sin32x  

W₂ = cos2x0-2sin2xsin2 2x=cos2x sin22x

Jadi,      U₁ = W1W=-sin32x2=12-sin32xdx

                                     =  12-sin22xsin(2x)

                                     = 12 1-cos22x(sin(2x))

                                     = sin x – sin 2x cos²2x

                                     = 12cos2x+ 16cos32x+ C

            U₂ = W2W=cos2x sin2 2x2

                        = 12cos2x sin22xdx                          Dik :     U = sin 2x

                                                                                    Du = 2 cos 2x dx

                                                                                    Cos 2x dx = 12du

                                                                                    = 12  12du.u2   

                                                                                    = 14u2du

                                                                                    = 14.13u3+C

                                                                                    = 112u3+ C

                                                                                    = 112 sin32x+C

            Yp        = U₁Y₁ + U₂Y₂

                                                = _{1 2cos2x+16cos32xcos2x+ 112sin32x.sin2x}

                                                = _{1 2cos2x+16cos42x+112sin42x}