Ecuaciones
Enviado por shaomy02 • 2 de Diciembre de 2011 • 424 Palabras (2 Páginas) • 487 Visitas
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
...