Ecuaciones diferenciales ejercicios de primer orden

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ECUACIONES DIFERENCIALES

PRESENTADO POR
CRISTIAN RAMIREZ OLARTE
CARLOS BOVEA ISAZA

CORPORACION UNIVERSITARIA DE LA COSTA
CUC
IV SEMESTRE DE INGENIERIA INDUSTRIAL
GRUPO AN
BARRANQUILLA - ATLANTICO

dy/dx=5y

dy/dx-5y=0

P_((x))=-5

e^∫▒〖-5dx〗=e^(-5x) → d/dx [e^(-5x).y]=e^(-5x) (0)

d/dx e^(-5x) y=0

e^(-5x) y=C

y=Ce^5x

dy/dx+2y=0
P_((x))=2e^(∫2dx)=e^2x → d/dx [e^2x y]=e^2x (0)
e^2x y=C
y=Ce^(-2x)

dy/dx+y=e^3x

P_((x) )=1

e^(∫dx)=e^x

d/dx [e^x y]=e^3x e^x→de^x y=e^4x dx

e^x y=1/4 e^4x+C

y=1/4 e^3x+Ce^(-x)
3 dy/dx+12y=4

dy/dx+4y=4/3

P_((x) )=4

e^(∫4dx)=e^4x

d/dx [e^4x y]=4/3 e^4x

e^4x y=1/3 e^4x+C

y=Ce^(-4x)+1/3

y^'+3x^2y=x^2

dy/dx+3x^2 y=x^2

P_((x))=3x^2
e^(∫3x^2 dx)=e^(3^(x^3/3) )=e^(x^3 )
d/dx [e^(x^3 ).y]=x^2 e^(x^3 )
d[e^(x^3 ) y]=x^2 e^(x^3 ) dx
u=x^3
du=3x^2dx
e^(x^3 ) y=1/3 e^(x^3 )+C
y=e^(x^3 )/(3e^(x^3 ) )+Ce^(-x^3 )
y=Ce^(-x^3 )+1/3

y^'+ 2xy= x^3

dy/dx+2xy= x^3

P(x)=2x

e^∫▒2xdx=e^(2∫▒xdx)=e^(x^2)

d/dx= (ye^(x^2 ) ) = x^3 e^(x^2 )

ye^(x^2 )= ∫▒〖x^3 e^(x^2 ) dx= 〗 ∫▒〖x^2 e^(x^2 ) x dx 〗

u=x^2 ; du=2xdx

dv=e^(x^2 ) x dx ; v=1/2 e^(x^2 )

ye^(x^2 )= (x^2 e^(x^2 ))/2-1/2 ∫▒〖e^(x^2 ) 2x dx〗

ye^(x^2 )= (x^2 e^(x^2 ))/2-1/2 e^(x^2 )+ C

y=x^2/2-1/2+Ce^(-x^2 )

x^2 y^'+xy=1

dy/dx+1/x y=1/x+1/x^2

P(x)=1/x 〖==>e〗^(∫▒dx/x )=e^lnx=x

d/dx [xy]=1/x x .1/x^2x==>d[xy]=dx/x

xy=1+lnx+C

y=(1+lnx+C)/x
y^'=2y+ x^2+ 5

dy/dx- 2y=5+x^2

P(x)= -2==> e^∫▒〖-2dx〗= e^(-2x)

d/dx [e^(-2x) y]=(5+x^2)e^(-2x)

e^(-2x) y= ∫▒〖5e^(-2x) dx+ 〗 ∫▒〖x^2 e^(-2x) 〗 dx= -5/2 ∫▒〖e^(-2x) (-2〗 dx)+ ∫▒〖x^2 e^(-2x) 〗 dx

e^(-2x) y= -5/2 e^(-2x) -1/2 e^(-2x) (x^2+ x+1/2)+ C

y=-5/2-1/2 (x^2+ x+1/2)+Ce^2x

y=-5/2 - 1/2 x^2- 1/2 x - 1/4+Ce^2x

y=-22/8-1/2 x^2- 1/2 x+ Ce^2x

y= -1/2 x^2- 1/2 x-11/4+ Ce^2x

x dy/dx- = x^2 senx

dy/dx-1/x y=xsenx
P(x)= -1/x==>e^∫▒〖-dx/x 〗= e^(-∫▒dx/x)= e^(-lnx)=e^(lnx^(-1) )=x^(-1)=1/x

d/dx [1/x y]=xsenx/x==>d[y/x]=senx dx

y/x= -cosx+C

y=Cx-xcosx

x dy/dx+ 2y=3 ==>dy/dx+2/x y = 3/x

P(x)=2/x==> e^∫▒〖2 dx/x 〗= e^2lnx= e^ln⁡〖x^2 〗 = x^2

d/dx [x^2 y]=3/x x^2 ==>d[x^2 y]=3x dx〖 x〗^2 y=(3x^2)/2+ C

y=3/2+ Cx^(-2)

x dy/dx+2y=3

dy/dx+4/x y=x^2-1
P_((x))=4/x-→e^(∫4dx/x= e^(4lnx )= e^lnx4 =〖 x〗^4 )
d/dx [x^4 y]=[x^2-1] x^4-→x^4 y=(x^6-x^4 )dx
x^4 y=x^7/7-x^5/5+C
y=x^3/7-x/5+C x^(-4)

(1+x) dy/dx-xy=(x+x^2)
(1+x) dy/dx-xy=x(1+x)dy/dx-(x/(1+x))y=x--→P_((x))=-x/(1+x)
e^(∫x/(1+x) dx)- ∫xdx/(1+x)--→ ∫(1-1/(1+x))dx
-∫dx+∫dx/(1+x)=-x+ln1+x
e^(-x+ln⁡〖1+x〗 )=e^(-x) e^(ln1+x)=(1+x) e^(-x)
d/dx [e^(-x) (1+x)y]=(1+x) e^(-x) x
e^(-x) (1+x)y=∫(1+x) e^(-x) xdx
= ∫xe^(-x) dx+∫e^(-x) x^2 dx
=-xe^(-x)+∫e^(-x) dx-x^2e^(-x)+2∫e^(-x) xdx

e^(-x) (1+x)y=∫xe^(-x) dx-x^2 e^(-x)+2∫e^(-x) xdx
=3∫xe^(-x) dx+x^2 e^(-x)
=3(-xe^(-x) )+3∫e^(-x) dx=3(-xe^(-x) )-3e^(-x)+C
e^(-x) (1+x)y=-3xe^(-x)-3e^(-x)+C
y=(-3xe^(-x))/((1+x) e^(-x) )-(3e^(-x))/((1+x) e^(-x) )+Ce^x
y=(-3x)/(1+x)-3/(1+x)+Ce^x
x^2y^'+x(x+2)y=e^x
dy/dx+((x+2)/x)y=e^x/x^2
P_((x) )=(x+2)/x=-→e^(∫((x+2)/x)dx)=e^(∫dx+2∫dx/x)=e^(x+2lnx)
= e^(x+lnx^2 )=e^x e^(lnx^2 )=x^2 e^x
d/dx [x^2 e^x y]=(e^x e^x x^2)/x^2 =e^2x
x^2 e^x y=∫e^2x dx=1/2 e^2x+C
y=1/2 e^2x/(e^x x^2...
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