ECUACIONES DIFERENCIALES/ANIQUILADOR/

1.y´´-2y^'=sin⁡x y_p=2/5 cos⁡x-1/5 sin⁡x

(D^2-2D)y=0

m^2-2m

m(m-2) m_(1=) 0 m_2=2

y_c=C_1+C_2 e^2x

(D^2+1) sin⁡x=0

D(D-2)(D^2+1)=(D^2+1) sin⁡x

m_(1=) 0 m_2=2 m_3,4=±i

y=C_1+C_2 e^2x+C_3 cos⁡x+C_4 sin⁡x

y_p=A cos⁡x+B sin⁡x

〖y´〗_p=-A sin⁡x+B cos⁡x

〖y´´〗_p=-A cos⁡x-B sin⁡x

-A cos⁡x-B sin⁡x-2(-A sin⁡x+B cos⁡x )=sin⁡x

-A cos⁡x-B sin⁡x+2Asin⁡x-2B cos⁡x=sin⁡x

(-B+2A) sin⁡x=sin⁡x

(-A-2B) cos⁡x=0

-B+2A=1

-2B-A=0

2B-42A=-2 -B+2(2/5)=1

-2B-A=0 -B+4/5=1

-5A=-2 -B=1-4/5

A=2/5 B=-1/5

y_p=2/5 cos⁡x-1/5 sin⁡x

2. y^'''+y^'=cos⁡x y_p=-1/2 xcos x

(D^3+D)y=0

m^3+m

m(m^2+1) m_(1=) 0 m_2,3=±i

y_c=C_1+C_2 cos⁡〖x+〗 C_3 sen x

(D^2+1) cos⁡x=0

D(D^2+1)(D^3+1)=(D^3+1) cos⁡x

m_(1=) 0 m_2,3,4,5=±i

y=C_1+C_2 cos⁡x+C_3 sen⁡x+C_4 x cos⁡x+C_5 x sen⁡x

y_p=Ax cos⁡x+Bx sen⁡x

〖y´〗_p=A[cos⁡x-xsen x]+B[x cos⁡〖x+sen x]〗

〖y´´〗_p=A[-sen x-x cos⁡〖x+sen x]〗+B[cos⁡x-xsen x+cos⁡x]

〖y'''〗_p=A[-cos x-cos⁡〖x+x senx-cos⁡x]〗+B[sen x-xcos x+sen x-sen x]

A[-cos x-cos⁡〖x+x senx-xcos⁡x]〗+B[sen x-xcos x+sen x-sen x]+Axcos x+Bx senx = cos⁡x

(B)

...