Ecuaciones

Cristian Quintero Castañeda 2009217058

Gustavo Vergel 2005211015

2. y^''+a^2 y=f(x) (D^2+a^2 )y=f(x) (D^2+a^2 )y=0

La ecuación auxiliar es m^2+a^2=0 m=±ai

y_h=c_1 cos⁡(ax)+c_2 sin⁡(ax)

w(y_1,y_2 )=[■(cos⁡(ax)&sin⁡(ax)@-asin⁡(ax)&acos⁡(ax))]=〖cos〗^2 (ax)+sin^2 (ax)=a

u_1^'=[■(0&sin⁡(ax)@f(x)&cos⁡(ax))]/(w(y_1,y_2))=-(f(x)sin⁡(ax))/a=-f(x)sin⁡(ax)

u_1=-∫▒〖f(x) sin⁡(ax)dx〗=-1/a ∫▒〖f(x) sin⁡(ax)dx 〗=-1/a ∫▒〖f(μ) sin⁡(aμ)dμ 〗

u_2^'=[■(cos⁡(ax)&0@-sin⁡(ax)&f(X))]/(w(y_1,y_2))=-(f(x)cos⁡(ax))/1=f(x)cos⁡(ax)

u_2=∫▒〖f(x) cos⁡(ax)dx〗=1/a ∫▒〖f(x) cos⁡(ax)dx 〗=1/a ∫▒〖f(μ) cos⁡(aμ)dμ 〗

y_p=-1/a cos⁡(ax) ∫_0^x▒〖f(μ) sin⁡(aμ)dμ〗+1/a sen(ax)∫_0^x▒〖f(μ) cos⁡(aμ)dμ〗

y_p=1/a ∫_0^x▒〖f(μ)(sin⁡(aμ)*cos⁡(ax)-〗 cos⁡(ax)*sin⁡(aμ))dμ

y_p=1/a ∫_0^x▒〖f(μ)(sin⁡〖(a(x-μ)))〗 dμ〗

Luego, la solución general es:

y=c_1 cos⁡(ax)+c_2 sin⁡(ax)+1/a ∫_0^x▒〖f(μ)(sin⁡〖(a(x-μ)))〗 dμ〗

Como y(0)=y^' (0)=0

0=ac_1 ; c_1=0

y=c_2 sin⁡(ax)+1/a ∫_0^x▒〖f(μ)(sin⁡〖(a(x-μ)))〗 dμ〗

y^'=c_2 cos⁡(ax)-1/a^2 ∫_0^x▒〖f(μ)(sin⁡〖(a(x-μ)))〗 dμ〗+1/a [f(μ)(sin⁡〖(a(x-μ)))〗 dμ] d/dμ

y=1/a ∫_0^x▒〖f(μ)(sin⁡〖(a(x-μ)))〗 dμ〗

3.

x^3 y^'''+xy^'-y=xlnx

sea t=lnx ; x= e^t

dy/dx=dy/dt*dt/dx ; dy/dx=dy/dt*1/x ; dy/dt=x dy/dt ; (d^2 y)/(dx^2 )=1/x*d/dx (dy/dt)-1/x^2 *dy/dt

(d^2 y)/(dx^2 )=1/x*(d^2 y)/(dt^2 ) (1/x)-1/x^2 dy/dt ; x^2 (d^2 y)/(dx^2 )=(d^2 y)/(dx^2 )-dy/dt

(d^3 y)/(dx^3 )=d/dx (1/x^2 )[(d^2 y)/(dt^2 )-dy/dt]+1/x^2 [(d^3 y)/〖dt〗^3 dt/dx-(d^2 y)/(dt^2 ) dt/dx]

(d^3 y)/(dx^3 )=-2/x^3 ((d^2 y)/(dt^2 )-dy/dt)+1/x^2 ((d^3 y)/(dt^3 ) 1/x-(d^2 y)/(dt^2 ) 1/x)

x^3 (d^3 y)/〖dx〗^3 =(d^3 y)/〖dt〗^3 -3 (d^2 y)/(dt^2 )+2 dy/dt

reemplazando en ; x^3 y^'''+xy^'-y=xlnx

(d^3 y)/(dt^3 )-3 (d^2 y)/(dt^2 )+2 dy/dt+dy/dt-y=e^t t

(D^3-3D^2+3D-1)y=te^t ; (D-1)^3 y=te^t ; (D-1)^3 y=0 ; m_1=m_2=m_3=1

y_c=(c_1+c_2 t+c_3 t^2 ) e^t

f(t)=te^t es anulado por (D-1)^2 ; (D-1)^2 f(t)=0 ; (m-1)^2=0 ; m_4=m_5=1

y_p=t^3 (A+Bt) e^t

y_p^'=3t^2 (A+Bt) e^t+Bt^3 e^t+t^3 (4+Bt)e^t

y_p^''=6t(A+Bt) e^t+6Bt^2 e^t+6t^2 (A+Bt) e^t+2Bt^3 e^t+t^3 (A+Bt)e^t

y_p^'''=6(A+Bt) e^t+6Bt^2 e^t+6t(A+Bt) e^t+12Bte^t+6Bt^2 e^t+12t(A+Bt) e^t+6Bt^2 e^t+6t^2 (A+Bt) e^t+6Bt^2 e^t+2〖Bt〗^3

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