Solucionario Mate 3
1 F.I.M.E. MECATRONICA EJERCICIOS 1.1 1. _ ORDEN d2y dy dy + 13 + x 2 = 2 dx dx dx 3 3
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1
F.I.M.E.
MECATRONICA
EJERCICIOS 1.1
1. _
ORDEN
d2y dy dy + 13 + x 2 = 2 dx dx dx
3
3
d3y d 3y dy + 18 3 = 8 x + 3 3) dx dx dx
5
d3y dy 4) 3 − 5 x = 8 dx dx d 2 y 5) 2 = dx
1
SI
2
1
NO
3
5
NO
3
2
SI
2
1
SI
3
6
NO
2
6
NO
3
1
SI
5
2
NO
2
3
NO
x−2
d 2 y d3y 6) 2 + 3 x = 5 3 dx dx
3
d 2y dy 7) 2 + 7 x = 81 + dx dx d3y 8) 3 = dx d 5y 9) 5 dx
1
3
3
LINEALIDAD
3
d3y dy = 3x + 5 y dx 3 dx 4
2)
GRADO
dy dx d 2 y 2 = 81 + 2 dx
d 2 y dy 10 ) 2 = 5 − dx dx
5
1
4
2
5
8 de julio de 2008
2
F.I.M.E.
MECATRONICA
EJERCICO 1.2 Determinar si la solución general es o no de la ecuación general dada. 1) y = c + cx 2
−1
y + xy ′ = x 4 ( y ′) 2
dy = 0 − cx − 2 dx c 2 − cx −1 + x(cx − 2 ) = x 4 (cx − 2 ) c 2 − cx −1 + cx −1 = x 4 (c 2 x − 4 ) c2 = c2 2) e
Cosx
Si es solucion dy Seny + SenxCosy = Senx dx
(1 − Cosy ) = 0
(
)
(
)
dy = e Cosx Seny + (1 − Cosy )e Cosx (− Senx ) dx dy = e Cosx Seny y ′ − (1 − Cosy )e Cosx (Senx ) dx (1 − Cosy )e Cosx (Senx ) y′ = − e Cosx Seny
(
)
(1 − Cosy )e (Senx ) + SenxCosy = Senx Seny − e Cosx Seny Senx = Senx Si es solucion Cosx
(
3)
)
y = 8x5 + 3x2 + c
d2y 2 − 6 = 160 x 3 dx
dy = 40 x 4 + 6 x dx d2y = 160 x 3 + 6 2 dx 160 x 3 + 6 − 6 = 160 x 3 160 x 3 = 160 x 3
4)
Si es solucion
y = c1 Sen3 x + c 2 Cos 3 x
d2y 2 + 9 y = 0 dx
dy = 3c1Cos 3 x − 3c 2 Sen3 x dx d2y = −9c1Cos 3 x − 9c 2 Sen3 x dx 2 − 9c1Cos 3 x − 9c 2 Sen3x + 9(c1Cos3 x + c 2 Sen3 x) = 0 0 = 0 Si es solucion
8 de julio de 2008
3
F.I.M.E.
5)
MECATRONICA
dy + y = e−x dx
y = ( x + c )e − x
dy = −e − x ( x + c ) + e − x dx − e − x ( x + c ) + e − x + −e − x ( x + c ) = e − x e−x = e−x 6)
Si es solucion dy − 5y = 0 dx
y = ce 5 x
dy = 5ce 5 x dx 5ce 5 x − 5ce 5 x = 0 0 = 0 Si es solucion 2
7)
d 2 y dy y 2 − = y 2 ln y dx dx
ln y = c1 Senx + c 2 Cosx
dy = y (c1Cos − c 2 Senx) dx d2y = y ( −c1 Senx − c 2 Cosx ) + (c1Cosx − c 2 Senx) dy dx 2 2 y[ y (c1Cos − c 2 Senx ) − (c1Cosx − c 2 Senx) dy ] − [ y (c1Cos − c 2 Senx) ] = y 2 ln y y 2 (c1Cos − c 2 Senx) + y 2 (c1 Senx − c 2 Cosx ) − y 2 (c1Cos + c 2 Senx ) = y 2 ln y y 2 (c1 Senx − c 2 Cosx ) = y 2 (c1 Senx − c 2 Cosx ) Si es solucion
EJERCICIOS 1.3 1) y = 7x² + 8x + C y = 14 x + 8 2) y = C1x² + C2 y' = C1(2x) y"= 2C1
2C1 = y' x
x( y”) = y'
8 de julio de 2008
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MECATRONICA
3) y = C1 sen 8x + C2 cos8x y' = 8C1 cos 8x – 8C2 sen 8x y" = -64 C1 sen 8x – 64C2 cos 8x y"= -64y y" + 64y = 0
4) y = tan (3x + c) Tan ־¹ (y) = 3x + c y' _ 1 + y² = 3 y' = 3 (1 + y² )
e 3 x ) + C2( e −5x ) 3x −5 x y' = 3C1( e ) – 5C2( e ) 3x −5 x ) + 25C2( e ) y" = 9C1( e
5) y = C1(
e 3 x ) + 5C2( e −5x ) 3x −5 x y' = 3C1( e ) – 5C2( e ) 3x -3 [ 5y + y' = 8C1( e )]
5y = 5C1(
e 3 x ) – 25C2( e −5x ) 3x −5 x ) + 25C2( e ) y" = 9C1( e 3x ) 5y' +y" = 24C1( e
e 3x ) 3x 5y' +y" = 24C1( e )
-15y - 3y' = - 24C1(
5y' = 15C1(
-15y - 3y' + 5y' +y" = 0
y" + 2y' – 15y = 0
6) y = x tan (x + c) Tan ־¹ (y/x) = x + c xy' – y x² __ 1 + y²_ x²
xy' – y x²
y² = 1 + x²
= 1 xy'- y = x² + y²
xy' = x² + y² + y
7) y = C1 senh(x) + C2 cosh(x) y' = C1 cosh(x) + C2 senh(x) y" = C1 senh(x) + C2 cosh(x) y"= y
8 de julio de 2008
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MECATRONICA
8) x²_ y²_ C1² = 1 - C2² - 2 C2²yy'
2C1²x_ C1
4
xC2² = -yy'(C1²) C2² = -C1²yy" - C1²(y')² C2² = C1²( -yy'' - (y')² )
=
C2
4
-yy' x = -yy'' - (y')²
xC2²_ C1² = -yy'
x [yy'' + (y')²] = yy'
9) y = x sen(x + c) y_ x = sen (x + c) sen ־¹ (y/x) = x + c (xy' - y)² = [
1−
y2 ]² x2
(x²)² xy' - y x²___
1−
= 1
x²(y')² - 2xy'y + y² = (1 - y²/x² ) x
4
y2 x2
x²(y')² - 2xy'y + y² + x²y² = x
4
10) ( x – C1)² + y² = C2² x² - 2xC1 – C1² + y² = C2² 2x – 2C1 + 2yy' = 0 [ 2 + 2yy''+ 2(y')² = 0 ] (1/2) 1 + yy" + (y')² = 0
11) Todas las líneas rectas Ax + By + C = 0 0 + By' = 0 By"= 0
(hay dos ctes. X y C )
y” = 0
y" = 0_ B
12) Todas las circunferencias de radio = 1 y centro en el eje x (x – h)² + (y – k)² = r² 2x – 2h + 2yy' = 1 x – h + yy' = 1 √(x – h)² = √(1- y²) x – h = √(1- y²)
r=1
h k (h , 0)
centro en el eje x
[√(1- y²)]² = (-yy')² 1 - y² = y²( y')² 1= y²( y')² + y² y²[(y')² + 1] = 1
8 de julio de 2008
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MECATRONICA
EJERCICIOS 2.1 1) 5x dx + 20y19dy=0 (5x5)/5 + (20y20)/20 = c x5+y20=c 2) 2sen2x dx + 3e3ydy = 2x dx -cos2x + e²y = x²+c e3y = cos 2x + x² + c 3) dr/ds = r ∫dr/r = ∫ds ln r = s + c eln r = es + ec r = ces 4) dx + dy + xdy = ydx dy(1+x) = dx(y-1) [1/(1+x)(y-1)] dy/(y-1) = dx/(1+x) = ln |y-1| - ln|1+x| = c e ln (y-1)/(1+x) = ec (y-1)/(1+x) = c y-1 = c(1+X) 5) x sen y dx + (x²+1) cos y dy = 0 [1/(seny)(x²+1)] ∫(x dx/x²+1) + ∫(cos y dy)/seny = ∫0 ½ ∫x2dx/x²+1 + ∫cot y dy = 0 ½ ln|x²+1| + ln|sen y| = c ln |√(x²+1) (sen y)| = c (x²+1)(sen²y) = c 7) dy/dx = (x+1)/(y +1) = ∫(y +1) dy = ∫(x+1) dy y5/5 + y = x²/2 + x + c [10] 2y5 + 10y = 5x² + 10x + c 2y(y +5) = 5x(x+2) + c 9) x sen xe-y dx – y dy =0 [1/ey] ∫x sen x dx -∫yey dy = 0 -x cos x + sen x – yey + ey = c
8) xy
dx + (y²+2) e-3x dy = 0
e3x dx + (y²+2)/y dy = 0 ∫e3x x dx + ∫dy/y² + ∫2dy/y = ∫0 (1/3 e3x x) + (1/9 e3x ) – (y-1 )– (2/3 y-3 )= c [9y³] y³( 3e3x + e3x ) – 9y² - 6 = 9cy³ e3x y³(3+1) - 9y² - 6 = cy³ 10) dy/dx = e3x + e2y
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MECATRONICA
dy/ e2y = e3x dx -1/2∫ e-2y 2dy = 1/3∫ e3x dx (-1/2)e-2y = (1/3)e3x +c [6] -3 e-2y = 2 e3x + c 11) dy/dx = (xy+3x-y-3)/(yx-2x+4y-8) =((y+3)(x-1))/((y-2(x+4)) dy (y-2)/y+3) = dx ((x+1)/(x+4) ∫( 1 - 5/(y+3) dy = ∫(1 - 5/(x+4) dx y – 5 ln|y+3| = x – 5 ln|x+4| + c e - (y+3)5 = ex – (x+4)5 + c
12) dy/dx = sen x (cos 2y - cos²y) dy/(cos 2y - cos²y) = sen x dx cos2y sen²y - cos²y = sen x dx -∫csc²y dy = ∫sen x dx cot y = -cos x + c
13) (x+1)dy + (y-1)dx = 0
y=3, x=0
∫dy/(y-1) + ∫dx/x+1) = ∫0 ln|y-1| + ln|x+1| = c ln(y-1)(x+1) = c [e^] (y-1)(x+1) = c c = (3-1)(0+1) c=2 (y-1) (x+1) = 2
14) x³dy + xy dx = x²dy + 2y dx
y=e , x=2
x³dy - x²dy = -xydx + 2ydx dy(x³-x²) = ydx (2-x) [1/(y x³-x²)] dy/y = (2-x/ x³-x²) dx = (2-x / x²(x-1)) A/x + B/x² + C/x-1 (x²(x-1)) = A(x²-x) + B(x-1) + Cx² Ax² + Ax + Bx – B + Cx² A=-1 B=-2 C=1 ln y = ∫1/(x-1) - ∫1/x - ∫2/x² ln y = ln|x-1| - ln|x| - 2/x ln y = ln (x-1)/x – 2/x + c c = ln|(x-1)/xy)| - 2/x [e^] c = (x-1)/xy – e2/x
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F.I.M.E.
MECATRONICA
EJERCICIOS 2.2
(
1) 6 x − 7 y
2
(6 x
)
2
)dx − 14 xydy = 0
− 7v 2 x 2 dx − 14 x 2 v 2 dx − 14 x 2 v 2 dx − 14 x 3 vdv = 0
2
6 x 2 dx − 21x 2 v 2 − 14 x 3 vdv = 0
[3x (2 − 7v )dx − 14 x vdv = 0] (2 − 7v1 )(x ) 2
2
3
2
3
3x 2 − 14 ∫ x 3 dx − ∫ 2 − 7v 2 dv = ∫ 0
(
)
1 − 14v du 3∫ dx + ∫ x 2 − 7v 2 − 14v u = 2 − 7v 2 du = −14vdv du = dv − 14v 3 ln x + ln 2 − 7v 2 = c
(
(
)
)
ln ( x ) 2 − 7v 2 = c 3
y2 3 ln ( x ) 2 − 7 2 x e ln 2 x
3
− 7 y 2x = c
= c
→ 2x3 = 7 y 2 x + c
8 de julio de 2008
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MECATRONICA
2) (2√ √xy − y)dx − xdy=0 y= vx dy= vdx + xdv (2√x²v − vx) dx − x(vdx + xdv) 2x√v dx − vxdx − xvdx − x²dv=0 [2x(√v −v) dx − x²dv=0] 1 ÷ x²(√v − v) 2∫ dx/x − ∫dv/(√v−v) = ∫0 √v(1−√v) u= 1−√v du= -1/2v-½ dv 2ln(x) + 2ln(1−√v)=c ℓ[ln(x²)(1−√v)²=c] (x²)(1−√v)²=c √( (x²)(1−√v)²)=√(c) X(1−√y/x)=c x − x √(x/y) =c x + c = √(xy)
y 3) y + xctg
dx − xdy = 0 x
y = vx dy = vdx + xdv
[vx + xctgv]dx − x(vdx + xdv) = 0 [vx + xctgv]dx − xvdx + x 2 dv = 0
[(xctgv)dx + x dv = 0] (x )(1ctgv) 2
2
1
1
∫ x dx + ∫ ctgv dv = 0 ln x + ln sec v = c x ... ln x sec = c y e y xSec = c x 4) (y+ √x²+y² ) dx- xdy=0 y=vx dy=vdy+ydv (vx+ √x²+v²x² )dx-x(vdx+xdv)=0 vxdx+ √x²(1+v²)dx- vxdx - x²dv=0 [x√(1+v²)dx - x²dv=0]1÷√(1+v²)(x²) ∫xdx/x² - ∫dv/√(1+v²) =∫0 u=x u²=v² a²=1 u=v a=1 du=dx du=dv ∫du/u - ∫du/√a²+u² =∫0 ln(x) – ln(v+√(1+v²)) = c ℓln(x/ v+√(1+v²))=ℓc
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MECATRONICA
(x/ v+√(1+v²))=c [ X=c(y/x+√x²+y²/x)]x x²=cy + √x²+y²
5)
dy 2 y 4 + x 4 = dx xy 3
y = vx dy = vdx + xdv
(xy )dy = (2 y 3
4
+ x 4 )dx
(
)
x(vx ) (vdx + xdv ) = 2(vx ) + x 4 dx 3
4
x 4 v 4 dx + x 5 v 3 dv = 2v 4 x 4 dx + x 4 dx 1 x 5 v 3 du = x 4 v 4 + 1 dx 5 4 x v +1
[
) ] (
(
)
v3 x4 = dv ∫ v 4 + 1 ∫ x 5 dx u = v4 +1 du = 4v 3 dv dv =
du 4v 3
[14 ln(v + 1) = ln x + c]4 4
4
ln v 4 + 1 = ln x + c v 4+1 =c x4 4
y +1 x =c x4 x4 + y4 =c x8 x 4 + y = cx 8
6) dy/dx = 2xyℓ(x/y)² x = vy dx = vdy+ydv y²dy + y²ℓ(x/y)² dy + 2x²ℓ(x/y)²dy − 2xyℓ(x/y)²=0 y²dy + y²ℓ(v)² dy + 2v²y²ℓ(v)²dy − 2v²y²ℓ(v)²dy − 2vy³ℓ(v)²dv=0 [y²(1+ ℓ(v)²)dy − 2vy³ ℓ(v)²dv=0] 1/(1+ℓv²)(y³) ∫dy/y − ∫2vℓ(v)²/ (1+ℓv²) dv =∫0 u=(1+ℓv²) du=2vℓv²dv −∫(2vℓ(v)²)(du/2vℓv²)/ (v/1) Ln(y) − ln(1+ℓv²)=c ℓ[ln(y/1+ℓv²)=c] (y/1+ℓv²)=c Y=c(1+ℓ(x²/y²))
8 de julio de 2008
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7)
MECATRONICA
dy 2 xy = 2 dx x − y 2
x = vy dx = vdy + ydv x 2 − y 2 dy = 2 xydy
(vy )2 dy − y 2 dy = 2(vy )y(vdy + ydv ) v 2 y 2 dy − y 2 dy = 2vy 2 (vdy + ydv ) v 2 y 2 dy − y 2 dy = 2v 2 y 2 dy + 2vy 3 dv
[(− 2vy dv) = y (v 3
− 2∫
2
2
) ] y (v1 + 1)
+ 1 dy
3
2
vdv y2 = dy v2 +1 ∫ y3
− ln v 2 + 1 = ln y + c
(
)
c = ln v 2 + 1 ( y )
(
)
c = v2 +1 y
c x2 y 2 = 2 + 1 y y 2 2 cy = x + y 8) (x²+y²)dx +(x² − xy)dy=0 x=vy dx=vdy+ydv x²dx+y²dx +x²dy − xydy=0 v²y²(vdy+ydv)+y²(vdy+ydv) +v²y²dy − vy²dy=0 v³y²dy+v²y³dv+y²vdy+y³dv +v²y²dy − vy²dy=0 [y² (v³ +v+v²−v)dy+y³(v²+1)dv=0]1/( v³+v²)(y³) ∫dy/y + ∫v²+v/ v³+v² dv =∫0 v²+1 / v²(v+1)=A/v + B/v² + C/v+1 v²+1 = A(v²)(v+1) + B (v+1) + C(v²) 2=C 1=B 2=2 A + 2 +2 2−4=2A -2/2=A A=-1 -∫1dv/v + ∫1dv/v² + 2∫dv/v+1 Ln(y)−ln(v)−v-¹ + 2ln(v+1)= c ℓ[ln(y)(v+1)²/v= c +1/v] y(v+1)²/v=cℓ(1/v) y(v²+2v+1)/v=cℓ(1/v) yv +2y +y/v =cℓ(1/v) x[x+2y+y²/x=cℓ(y/x) x²+2xy+y²=cℓ(y/x) (x+y)²= =cℓ(y/x)
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9)
MECATRONICA
(
)
(
)
y x 2 + xy − 2 y 2 dx + xx 3 y 2 − xy − x 2 dy = 0
x 2 ydx + xy 2 dx − 2 y 3 dx + 3 xy 2 dy − x 2 ydy − x 3 dy = 0 y = vx dy = vdx + xdv
(
)
(
)
x 2 (vx )dx + x v 2 x 2 dx − 2v 3 x 3 dx + 3x v 2 x 2 (vdx + xdv ) − x 2 ( xv )(vdx + xdv ) − x 2 (vdx + xdv ) = 0 x vdx + x v dx − 2v x dx + 3 x v (vdx + xdv ) − x 3 v (vdx + xdv ) − x 3 vdx + x 4 dv = 0 3
3
2
3
3
2
2
x 3 vdx + x 3 v 2 dx − 2v 3 x 3 dx + 3x 3 v 3 dx + 3x 4 v 2 − x 3 v 2 dx + x 4 vdx − x 3 vdv + x 4 dv = 0 x 3 v 3 dx + 3 x 4 v 2 dv + x 4 vdv + x 4 dv = 0 1 v 3 x 3 dx + x 4 3v 2 + v + 1 dv = 0 3 4 v x
[
(
(
)
)
] ( )
x3 3v 2 + v + 1 dv dx + =0 4 x v3 1 3dv dv dv ∫ xdx + ∫ v + ∫ v 2 + ∫ v 3 = 0 v −2 ln x + 3 ln v − v −1 − =c 2 1 1 3 2 ln x (v ) − v − 2v 2 = c − 2v − 2v ln xv 3 + 2v + 1 = v 2 c 2 y2 y3 y y (c ) x 2 2 ln 2 1 − + + = 2 2 x x x x
3 − 2 y 2 ln y 2 + 2 xy + x 2 = y 2 c x
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MECATRONICA
10) (x²y+y³)dx −2x³dy=0 : y=3 cuando x=2 y=vx dy=vdx+xdv x²ydx+y³dx −2x³dy=0 vx³dx+v³x³dx −2x³vdx− 2x dv=0 [x³(v+v³−2v)dx −2x dv=0]1/(v³−v)x ∫dx/x −2∫dv/(v³−v) =∫0 (v³−v)=v(v²−1)=v(v+1)(v−1) 1/v(v+1)(v−1)=A/v +B/v+1 + C/v−1 1= A(v+1)(v−1) + B(v)(v−1) + C(v)(v+1) 1= A(1)(-1) A=-1 1=C(2) C=1/2 1=B(-1)(-2) 1=2B B=1/2 ∫dx/x−2[-∫dv/v +1/2∫dv/v+1 + 1/2∫dv/v−1]=0 Ln(x)−2[-ln(v) +1/2ln(v+1) +1/2ln(v-1)]=c Ln(x)+2ln(v) −ln(v+1) − ln(v−1)=c ℓ[ln(xv²/(v²−1) =c (xv²/(v²−1) =c xv²=c(v²−1) x(y²/x²)=c(y²/x²−1) 9/2=c(9/4−4/4) 9/2=c(5/4) 36/10=c 18/5=c (y²/x)=18/5(y²/x²−1) X[5y²=18x(y²/x²−1)] 5y²x=18y²−18x² 5y²x=18(y²−x²)
11)
y y x x + x ye dx − xe dy = 0
y = vx dy = vdx + xdv xdx + vxe v dx − xe v (vdx + xdv ) = 0 xdx + vxe v dx − xve v dx − x 2 e v dv = 0 1 xdx − x 2 e v dv = 0 2 x 1 v ∫ x dx − ∫ e dv = ∫ 0 ln x − e v = c
[
]
ln x − e
ln 1 − e
y x
0 1
=c
=c
c = −1 ln x − e
y x
= −1 →→→ ln x = e
y x
−1
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12.- y²dx + (x²+xy+y²)dy=0
MECATRONICA y=1 cuando x=0
X=vy dx=vdy+ydv y²dx + x²dy + xydy + y²dy=0 vy²dx + y³dv + v²y²dy + vy²dy + y²dy=0 [y²(v²+2v+1)dy +y³dv=0]1/(v²+2v+1)(y³) ∫dy/y + ∫dv/(v²+2v+1) =∫0 (v²+2v+1)=(v+1)²= (v+1)-² Ln(y)− (v+1)-¹=c Ln(1)−1/(0/1 +1)=c c=-1 y[x+y/yln(y)−1=c(x+y/y)] (x+y)ln(y) +x=0
EJERCICIOS 2.3 1) + + + + − =
= + 2 = + 2 = − 2
+ 3 − 2 + 2 − 1 = 0 + 3 + −2 − 6 + 2 − 1 = 0 + 3 + −2 − 6 + 2 − 1 = 0 + 3 − = 0 + 3 − 7 = 0 2 + 3 − 7 = 0 2 2 + 6 − 14 = + 2 + 6x + 2y − 14y = c + 2 + 6x + 12y − 14y = c + 2 + 6x − 2y = c
2) + − + + + =
=+ = + = −
− 3 − + + 4 = 0 − − 3 + 3 + 2 + 4 = 0 − 3 + 7 = 0 − 3 + 7 = 0
− 3 + 7 = " 2 − 6 + 14 = " + − 6 + + 14 = " + 2 + − 6 − 6 + 14 = " + 2 + + 8 = " + 6
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MECATRONICA
3) − + + − $% − = & = 2 − 3 = 2 − 3 3 = 2 −
+ 2 + 2 − 12 − = 0 + 2 + 4 − 2 − 2 + = 0 + 2 + 4 − 2 + −2 + 1 = 0 1 '5 + −2 + 1 = 0) 5 −2 + 1 + = 0 + * 5 +
1 −2 + = 0 5
1 2 − + ln = 5 5
1 2 . − + ln = / 5 5 5 5 − 2 + ln = 5 − 4 + 6 + ln2 − 3 = + 6 + ln2 − 3 =
4) + − + − =
= 2 + = 2 + − = 2
− * + − 2 − 1 = 0 2 − + − 2 − 1 = 0/ 2 . * 2 − − 22 − 1 = 0 − − 4 + 2 = 0 − 5 − 2 = 0 1 ' − 5 − 2 = 0) * + 5 − 1 − = 0 5 − 1
− = 0 5 − 1 1 2 = + 5 − 1 5 55 − 2 1 2 + − = 0 * + 5 55 − 2 2 1 + − = 0 5 5 − 2 5
2 1 + ln05 − 20 − = " 25 5 5 + 2 ln05 − 20 − 25 = " 52 + + 2 ln052 + − 20 − 25 = "
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MECATRONICA
10 + 5 + 2 ln010 + 5 − 20 − 25 = " 10 − 20 + 2 ln010 + 5 − 20 = " 5 − 10 + ln010 + 5 − 20 = "
5) − + 1 − ' − + 2) = = − 2 = − 2 = 2 +
+ 52 + − 2 + 9 = 0 2 + + 10 + 5 − 2 − 9 = 0 + 5 + 2 + 10 − 2 − 9 = 0 + 5 + = 0 + 5 + = 0 + 5 + = 05 2 2 + 10 + 2 = 0 − 2 + 10x − 2y + 2y = 0 4
6) + + = = 2 + 3 = 2 + 3 − 3 = 2
− 3 + + = 0 2* 2 − 3 .2 * + + = 0/ 2 2 − 3 + = 0 + − 3 = 0 1 ' + − 3 = 0) * + −3 + = 0 −3 + = 0 −3 ln0 − 30 + = " ln0 − 30 = " − 6 7809:;0 = 6 " = 3 − + 7
−
6 > " + = 3 + 7
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9) + − = 1 − − 2 + 3 − 1 = −−5 + 2 + 3
= 2 + 3 = 2 + 3 − 3 = 2
− 3 + = −−5 + 2 − 3 − + 3 . = 5 − / 2 2 − 3 − + 3 − 10 + 2 = 0 − 1 + −3 + 3 − 10 + 2 = 0 1 ' − 1 − + 7 = 0) +7 +1 − = 0 +7 − 1 *
1 + 7B − 1 − − 7 -8 *−1 −
8 + − = 0 +7
− 8 ln + 7 − =
'2 + 3 − 8 ln2 + 3 + 7 − = 0) − − 4 ln2 + 3 + 7 =
1 2
10) + + + + − =
C = C = 2 C = D = D = 2 D = 2
C D + + C + D − 3 * + = 0 2 2 C D .C + D + 4 * + + C + D − 3 * + = 0/ 2 2 2 C + D + 4D + C + D − 3C = 0
C + D + 4 *
=C+D = C + D D = − C
+ 4 − C + − 3C = 0 − C + 4 − 4C + C − 3C = 0 + 4 − 7C = 0
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+ 4 − 7C = 0
+ 4 − 7C = " 2 4 + 4 − 7C = "5 2 2 + 8 − 14C = " C + D + 8C + D − 14C = " C + 2CD + D + 8C = " + 6C
E + 2 + E + 8 = " + 6 11) − F − + − + = '62 − 3 − 3) + 2 − 3 + 2 = 0
D = 3 D = 6 GI = J
C = 2 C = 4 GH = E
C D + C − D + 2 = 0/ 24 4 6 '6C − D − 3)4D + C − D + 26C = 0 '24C − D − 12)D + C − D + 26C = 0
.6C − D − 3
=C−D = C − D C = + D
24 − 12D + + 26 + D = 0 24D − 12D + 6 + 6D + 12 + 12D = 0 24 − 12 + 6 + 12D + 6 + 12 = 0 1 '30D + 6 + 12 = 0) 30 6 + 12 = 0 D + 30 D +
6 12 + = 0 30 30
12 6 + ln = 30 30 12 6 ln = / 5 .D + + 30 30 5D + + 2 ln = 53 + C − D + 2 ln = 15 + 2 − 3 + 2 lnC − D = D+
'12 + 2 + 2 ln2 − 3 = ) 6 + + ln2 − 3 =
1 2
8 de julio de 2008
20
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EJERCICIO 2.5 Encontrar las soluciones de las siguientes ecuaciones diferenciales. 1) ( 2 xCosy + 3 x y )dx + ( x − x Seny − y )dy = 0 2
3
2
∂M = 2 xSeny ∂y
∫ 2 xCosydx + ∫ 3x
2
∂ = 2 xSeny ∂x
∫ x dy + ∫ x 3
ydx
x 2 Cosy + x 3 y + c
y2 2
y2 =c 2
(x + Seny) dx + (xCosy - 2y) dy
∂M = Cosy ∂y
∂ = Cosy ∂x
∫ xdx + ∫ Senydx
∫ xCosydy − ∫ 2 ydy
x2 + xSeny + f ( y ) 2 f ( xy) :
3)
Senydy − ∫ ydy
x 3 y − x 2 Cosy − f ( xy) : x 2 Cosy + x 3 y −
2)
2
xSeny − y 2 + f ( x) x2 + xSeny − y 2 = c 2
dy 2 + ye ( xy ) = dx 2 y − xe ( xy )
∫ 2dx + ∫ ye
( xy )
∫ 2 ydy − ∫ xe
dx
( xy )
dy
− y 2 − e ( xy ) + f ( x)
2 x + e ( xy ) + f ( y )
f ( xy ) : e ( xy ) + 2 x − y 2 = c
4)
dy − 3 xy 2 = dx x 3 + 2 y 4
∂M = 3x 2 ∂y
∫ 3xy dx 3
∂ = 3x 2 ∂x
∫ x dy − ∫ 2 y 3
yx 3 + f ( y )
yx 3 + f ( xy ) : yx 3 +
4
dy
2y5 + f ( x) 5
2 y5 =c 5 8 de julio de 2008
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5) 2 xydx + (1 + x )dy = 0 2
∂M = 2x ∂y
∂ = 2x ∂x
∫ 2 xydx
∫ dy − ∫ x
x 2 y + f ( y)
2
dy
y + x 2 y + f ( x) f ( xy ) : x 2 y + y = c
EJERCICIOS
2.6
1) >xdy + ydx = x 4 y 8 dy 1 d ( xy ) = x y dy xy d ( xy ) x 4 y 8 dy = 4 4 ( xy ) ( xy ) 4
∫ ( xy )
−4
4
8
d ( xy ) = ∫ y 4 dy
( xy )−3 y 5 3 = + c − 15 ( xy ) −3 5
5 + 3 x3 y 8 = C ( xy )
3
2) >3 ydx = xy 3 dy − 5 xdy 3 ydx + 5 xdy = xy 3dy x 2 y 4 3 5 3 7 1 d ( x y ) = x y x3 y5 d ( x3 y5 )
∫ ( x y ) = ∫ y dy 3
2
5
y3 3 5 x y ln = ) 3 + c 3 (
3ln ( x3 y 5 ) = y 3 + c ln ( x 3 y 5 ) = y 3 + c 3
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3) >.xdy − ydx = xy 5 dy 1 xdy − ydx = xy 5 dy xy dy dx 4 ∫ y − ∫ x = ∫ y dy ln y − ln x =
y5 +c 5 5
y y5 ln x = 5 + c y 5ln e = y 5 + ce x 5 y 5 5 5 = y + c x x 5
y 5 = e y cx 5
dy 3 yx 2 → ( x3 + 2 y 4 ) dy = ( 3 yx 2 ) dx 4) > = 3 4 dx x + 2 y 1 x3 dy + 2 y 4 dy = 3 yx 2 dx 2 x −2 4 xdy + 2 x y dy = 3 ydx 3 ydx − xdy = 2 x −2 y 4 dy ( x2 y −2 )
∫d (x
3
y −1 ) = 2∫ y 2 dy
3 −1 2 3 x y = 3 y + c 3 y 3x3 = 2 y 4 + cy 3x3 = y ( 2 y 3 + c )
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5) >( y − xy 2 ) dx + ( x + x 2 y 2 ) dy = 0
ydx − xy 2 dx + xdy + x 2 y 2 dy = 0 ydx + xdy − xy 2 dx + x 2 y 2 dy = 0 d ( xy ) − xy 2 dx + x 2 y 2 dy = 0
( xy )
−2
∫ ( xy )
d ( xy ) −
−2
− ( xy
xy 2
( xy )
d ( xy ) − ∫
2
dx +
1
( xy )
2
x2 y2 dy = 0 ( xy )
dx + dy = ∫ 0 x ∫
) d ( xy ) − ln x + y = c − ( xy ) d ( xy ) − ln x + y = c − xy −1
−1
1 + xy ln x − xy 2 = cxy 1 + xy ln x = cxy + xy 2 1 + xy ln x = xy ( c + y )
dy xy 2 − y 6) > = → xdy = xy 2 − ydx dx x xdy = xy 2 dx − ydx 1 xdy + ydx = xy 2 dx 2 ( xy ) d ( xy )
∫ ( xy )
2
d ( xy )
∫ ( xy )
2
( xy ) dx 2
=∫ =∫
( xy )
2
dx x
( xy )−1 = ln x + c − xy 1 = − xy ln x + xyc 1 + xy ln x = cxy
8 de julio de 2008
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MECATRONICA
7) >( x 3 y 2 + x ) dy + ( x 2 y 3 − y ) dx xdy − ydx + x3 y 2 dy + x 2 y 3 dx = 0 xdy − ydx = − x 3 y 2 dy − x 2 y 3 dx
1
( xy )
2
dy dx 2 − 2 = − xdy − ydx x y xy dx dy xdy + ydx = 2 − 2 xy x y xy dx dy ∫ ( xy )d ( xy ) = ∫ x − ∫ y dx dy ∫ ( xy )d ( xy ) − ∫ x + ∫ y = c
( xy )
2
2
+ ln y − ln x = c
( xy )2 y + ln = c 2 x 2
( xy )
2
2
y + ln = c x
8) >4 ydx + xdy = xy 2 dx 1 d ( x 4 y ) = x 4 y 2 dx 4 2 (x y)
∫
d ( x4 y )
( x y) 4
2
=∫
dx x4
−1 x −3 4 − = − + c − 3x 4 y x y ( ) 3 4 3 = xy + cx y
3 = y ( x + cx 4 )
8 de julio de 2008
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9) >( y 2 − y ) dx + xdy = 0 y 2 dx − ydx + xdy = 0 1 xdy − ydx = − y 2 dx 2 y −x = − ∫ dx y
∫ d
−x y = −x + c ( − y ) x = xy + cy x = y ( x + c)
10) >ydx − xdy = ( x 2 + y 2 )
2
( xdx + ydy )
1 ydx − xdy = ( x 2 + y 2 )2 ( xdx + ydy ) ( x 2 + y 2 ) ydx − xdy = ( x 2 + y 2 ) ( xdx + ydy ) 2 2 x y + ( )
x 1 2 2 = ∫ ( x + y ) 2 ( xdx + ydy ) y 2 2 2 2 ( x + y ) + c 4 tan −1 x = 1 2 y 2
∫ d tan
−1
2 x 4 tan −1 = ( x 2 + y 2 ) + c y
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11) >xdy − ydx = ( x 2 + xy − 2 y 2 ) dx 1 xdy − ydx = x 2 dx + xydx − 2 y 2 dx 2 x + xy − 2 y 2 1 xdy − ydx x2 = dx 2 1 ( x + xy − 2 y 2 ) x2 y d dn dn x ∫ y y 2 = ∫ dx ⇒ 1 + n − 2n2 = (1 + 2n )(1 + n ) 1+ − 2 x x dn A B ∫ (1 + 2n )(1 − n ) = (1 + 2n ) + (1 − n )
( 2 ) dn ( −1) dn 2 1 + ∫ ∫ ( 2 ) 3 1 + 2n ( −1) 3 1 − n 1 1 1 1 + 2n ln 1 + 2n − ln 1 − n ⇒ ln 3 3 3 1− n A − An + B + 2 Bn = 1 A+ B =1 2B − A = 0 3B = 1 1 2 B = ,A= 3 3
( ) = x ⇒ 1 ln x +x2 y = 1 ln x + 2 y = x + c 3 ( ) 3 x −x y 3 x − y
1 + 2 y x 1 ln 3 1− y x
x + 2y ln = 3x + c x− y
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EJERCICIOS 2.7
dy 1) > − 3 y = 6 dx P = −3
e∫
e =e u
dy −3 x dx − 3 y = 6 e dx dy −3 x −3 x dx − 3 y ( e ) = 6e
∫ d ( ye
−3 x
P ( x ) dx
dx
−3 x
u = −3 x; du = −3 − 3 dx e ∫ = e − 3 x dx
) = ∫ 6 e −3 x
6 ye −3 x = − e −3 x 3 −3 x ye = −2e −3 x + c ye −3 x + 2e −3 x = c e −3 x ( y + 2) = c
dy 2) > + y = senx dx F.I = e ∫dx = ex
2 ∫ e x senxdx = − e xcos x + e x senx
∫ d ( ye x) = ∫ e x senxdx
∫e
ye x = ∫ e x senxdx
C
dv = senxdx u = ex du = exdx v = − cos x
[ yex =
uv − ∫ vdu
− e x cos x − ∫ − cos x ex
x
senxdx =
ex
1 ( − e xcos x + e x senx) + 2
1 ( − e xcos x + e x senx) + C ] 2 / 2
2y = − cos x + senx + ce -x
− e x cos x + ∫ cos x exdx dv = cos xdx u = ex du = exdx v = senx uv − ∫ vdu
e x senx − ∫ senx exdx
∫e
x
senxdx = − e xcos x + e x senx − ∫ senx ex
∫e
x
senxdx + ∫ e x senxdx = − e xcos x + e x senx
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3) >y1 − 2 xy = x
x2
P = −2x
dy − 2 xy = x dx − x2 − x2 ∫ d ye = ∫ xe
(
−2 2 −2 xdx = e 2 dx = e− x F .I = e∫
)
u = −x du = −2 xdx 2
1 − x2 e ( −2 ) xdx 2∫ 1 − x2 − x2 2 ye = − e + c − x 2 2 e c 2 y = −1 + − x 2 e 2
ye − x = −
2 y + 1 = ce x
2
4) >y´−7 y = sen ( 2 x )
P = −7
dy − 7 y = sen 2 x dx
∫ d ( ye ) = ∫ sen2 xe −7 x
u = e−7 x du = −7e−7 x dx
−7
dx
u = sen 2 x du = − cos 2 x
F .I = e ∫
−7 dx
= e −7 x
∫ udv =uv − ∫ vdu dv − ∫ sen 2 xdx 1 v = − cos 2 x 2
1 7 sen 2 x = − cos 2 xe −7 x − ∫ cos 2 xe −7 x dx 2 2 dv = ∫ cos 2 x u = e−7 x 1 du = 7e−7 x x = sem2 x 2 1 7 1 1 −7 −7 x −7 x −7 x ∫ e sen2 x = − 2 cos 2 xe − 2 2 sen2 xe − ∫ 2 sen2 x − 7e dx
∫e
−7
4 −7 1 7 49 e sen 2 xdx = − cos 2 xe −7 x − sen 2 xe−7 x − ∫ sen 2 xe−7 x dx 4∫ 2 4 4 53 −7 1 7 e sen 2 x = − cos 2 xe−7 x − sen 2 xe −7 x 4 ∫ 2 4 2 7 −7 −7 x −7 x ∫ sen2 xe dx = − 53 cos 2 xe − 53 sen 2 xe 2 7 −7 x −7 x −7 x 53 ye = − 53 cos 2 xe − 53 sen 2 xe e −7 x c 53 y = −2 cos 2 x − 7 sen 2 x + −7 x e 53 y + 2 cos 2 x + 7 sen 2 x = ce7 x
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dy 5) >x − 4 y = x 6 e x dx dy 6 x 1 x dx − 4 y = x e x dy 4 y − = x5e x dx x
P=−
4 x
4
F .I . = e
∫ − x dx
dx = e
−4
dx
∫x
−4
= e −4ln x = e ln x = x −4
dy 4 y = x5e x x −4 − dx x
∫ d ( yx ) = ∫ xe −4
xdx
u=x
dv = ∫ e x v = ex
du = dx yx −4 = xe x − ∫ e x dx yx −4 = xe x − e x + c
1 − xe x = c − e x ) −4 x x e y x c 1 − −4 = −4 x − −4 x e x x e x 4 y cx 5 4 x x − x = x − x (e ) e e
( yx
−4
y − x 5e x = cx 4 − x 4 e x y + x 4 e x = cx 4 + x5 e x y + x 4 e x = x 4 ( c + xe x )
6) >( x 2 + 9 )
dy + xy = 0 dx dy 2 1 ( x + 9 ) dx + xy = 0 x 2 + 9 dy xy + 2 =0 dx x + 9 u = x2 du = 2 xdx
∫d (y
x
x P= 2 x +9
F .I . = e e
ln
∫ x2 +9
( x +9) 2
1
dx = e 2
2x
∫ x2 +9
1
= e2
(
ln x 2 + 9
)
= x2 + 9
)
x2 + 9 = ∫ 0
y x2 + 9 = c
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dy + ysenx = 1 a 2 + b 2 dx dy 1 cos x dx + ysenx = 1 cos x 1 dy senx +y = cos cos x dx dy dx + y tan x = sec x P = tan x
7) >cos x
e∫
tan x
= eln sec x = sec xdx
dy dx + y tan x = sec x ( sec xdx ) dy 2 dx + y tan x ( sec xdx ) = sec xdx
∫ d ( y sec x) = ∫ sec x xdx 2
( y sec x) = tan x + c tan x c y= + sec x sec x y = senx + cos( x)c
cos 2 xsenxdy + ( y cos 3 x − 1)dx = 0 [ cos 2 xsenxdy + ( y cos 3 x − 1) dx = 0 ] 1/ dx 8)
[(cos2 x senx)
dy + (y cos3 x − 1) = 0 ] 1 / (cos2 x sen x) dx
dy + (y cos3 x − 1) / (cos2 x senx) = 0 dx dy + (y cos3 x − 1) / (cos2 x senx) = 1 / (cos2 x senx) dx = e∫cos x/sen xdx = e∫du / u = eln sen x = sen x u = sen x du = cos x
∫ senx (1 / (cos x senx) dx y sen x = ∫ 1 / cos x dx y sen x = ∫ (1 / cos x) dx y sen x = ∫ sec x dx
y sen x =
2
2
2
2
ysenx = − tan x + c
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dy 1 − e −2 x dx 9) > + y = x − x F .I = e ∫ = e x dx e +e P =1 dy 1 − e −2 x x + y = x − x e dx e +e dx x ∫ d ( ye ) = ∫
e x dx − e − x dx e x + e− x
u = e x + e− x du = e x dx − e − x dx ye x = ∫
du u
ye x = ln ( e x + e − x ) + c
dy 10) > + y tan x = cos 2 x; y = −1 , cuando x = 0 dx −1 tan xdx −1 P = tan x F .I = e ∫ = e − ln ( cos x ) = e ln ( cos x ) = ( cos x ) dx −1 dy 2 + y tan x = cos x ( cos x ) dx dx cos 2 x 2 d y x = cos ( ) ∫ ∫ cos x dx
(
)
y ( cos x ) = ∫ cos xdx −1
−1 dy 2 + y tan x = cos x ( cos x ) dx dx cos 2 x 2 d y x cos = ( ) ∫ ∫ cos x dx
(
)
y ( cos x ) = ∫ cos xdx −1 −1
y ( cos x ) = senx y = [ senx + c ] cos x −1 = [ sen0 + c ] cos 0 = c = −1 y = senx cos x − 1cos x
8 de julio de 2008
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MECATRONICA
dy + y = ln ( x ) dx
11) >( x + 1)
y = 10 cuando x = 1
dv = ∫ dx v=x y ( x + 1) = x ln x − x + c 10(2) = 1ln1 − 1 + c 10(2) = 0 − 1 + c = c = 21 y ( x + 1) = x ln( x) − x + 21
EJERCICIOS
2.8
+ = =2 = 1: = 1:2 = :1 :1 = = −:2
−:2 + :1 = :1 2
1 1 .− 2 + = 2 / − 2 − =−
:2 2 :2 2
=−
= K: =
:2 2
:2
= − 2 + =
:2 :1 2
:2 2
=
+
:2 2
− =
+
:
− 2 = 63 2 :2
2
= −2
= K :2 = :
2
=: 2 = 2 2 =2 + 2
2 + − 2 = 63 3 63 2 = 2
2
2
2 2 :22
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MECATRONICA
62 = 3 + 3
:2
6 = + 3 −2 :2 3 = +4 2 = :2 2 + 43 − = ;
=
=
M122 2 − N =
=1
= =: = −: + :
=K =
122 2 :2 − : = −: + : 122 = 1 O122 = P 2
12 = −:1 + 12 = −:1 : + 120 = −1:1 :0 + = 13
12 = −:1 : + 13
EJERCICIO 3.1
1) (D² + 9D + 18)y = 0 m² + 9m + 18 = 0
m= -6 y= C1
m = -3
e −6 x + C2 e −3 x
(m + 6)(m+3) = 0
2) (D² - 3D – 10)y = 0
m= 5
m= -2
m² - 3m – 10 = 0
y= C1 e
5x
+ C2
e −2 x
(m – 5)(m + 2)
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3) (D² + 4D – 21)y = 0
m= -7
m² + 4m – 21 = 0
y= C1 e
(m +7)(m – 3)
4)
5)
6)
m= 3 −7 x
+ C2
e3x
35
F.I.M.E.
MECATRONICA
7. (D³ - D² - D +1)y = 0 m³ - m² - m - 1 = 0 1| +1 -1 -1 +1 +1 0 -1 1 0 -1 0 m=1 m² - 1 = 0 m² = 1 m = ±1 m3 = -1 1 m1 = 1 m2 = 1 8)
( D 6 − 8 D 4 + 16 D 2 ) y = 0 D 2 ( D 4 − 8 D 2 + 16) y = 0 m 2 (m 4 − 8m 2 + 16) = 0 m1 = 0 m2 = 0 ( m 2 − 4) 2 = 0 m2 − 4 = 0 m2 = 4 m2 = 4 m= +−2 m3 = 2 m 4 = −2 m5 = 2 m 6 = −2 y = C1 + C 2 x + C 3 e 2 x + C 4 e 2 x + C5 e − 2 x + C 6 e − 2 x y = C1 + C 2 x + (C 3 + C 4 )e 2 x + (C 5 + C 6 )e − 2 x 9)
y = c1ex + c2x0ex + c3xe-x y = (c1 + c2x)ex + c3e-x
36
F.I.M.E.
MECATRONICA
10)
(D
2
)
3
− 6D + 9 y = 0
[(D − 3 ) ] y = 0 [(m − 3 ) ] = 0 2 3
2 3
m−3= 0 m =3
[
y = C1 + C 2 x + C 3 x 2 + C 4 x 3 + C 5 X
]
12 )( D 2 + 2 D + 10 )( D 2 − 2 D + 2) y = 0 ( m 2 + 2 m + 10 )( m 2 − 2 m + 2) = 0 a =1
a =1
b=2
b = −2
c = 10
c=2
− b ± b 2 − 4 ac 2a − 2 ± 2 2 − 4(1)(10 ) 2(1) − 2 ± 6i 2 − 1 ± 3i
− ( −2) ± ( −2) 2 − 4(1)( 2) 2(1) 2 ± 2i 2 1± i
y = (c1 sen 3 x + c2 cos 3 x )e − x + (c3 senx + c4 cos x )e x
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13. − ( D 4 + 8 D 2 + 16)Y = 0 m 4 + 8m 2 + 16 = 0 ( m 2 + 4) 2 = 0 m 2 = −4 m2 = − 4 m = ±2i a + bi a=0 b=2 y = (c1 + c2 x) cos 2 x + (c3 + c4 x) sen 2 x 14.- (D⁶+6D⁴+9D²)y=0 m⁶+6m⁴+9m²=0
m²(m⁴+6m²+9)=0 (m²+3)(m²+3)
√m²=√-3 √m²=√-3 (m₃=√3ί) (m₄=√3ί) (m₅=√3ί) (m₆=√3ί)
y=C₁+C₂x+(C₃+C₄x)Sen√3x +(C₅+C₆x)Cos√3x
15.-
( D 2 + 16 ) 2 y = 0 ( m 2 + 16 ) 2 = 0 m = − 16 m = ±4i
y = (C1 + C 2 ) sen 4 x + (C 3 + C 4 ) cos 4 x
16)( D 2 − 4 D + 5) 2 y = 0 (m 2 − 4m + 5) 2 = 0 m 2 − 4m + 5 = 0 − (−4) + − (−4) 2 − 4(1)(5) 2 4 + − 16 − 20 2 4 + −2i 2 2−i 2+i a=2 b =1
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F.I.M.E. 17. (D⁴⁴ - 8D³ + 32D² - 64D + 64)y = 0 m⁴ - 8m³ + 32m² - 64m + 64 = 0 √(m² - 4m + 8)² =√ 0 m² - 4m + 8 = 0 a=1 b=-4 c=8 4±√(-4² - 4(8)) /2 4±√(-16) /2 = 2±√(-16) = -2×2±√4 -4±2 = ±2 y = [(c1 + c2x)cos 2x + (c3 + c4x)sen2x] e2x
18)
( D 5 − D 4 − 2 D 3 + 2 D 2 + D − 1) y = 0 1 _/ 1 − 1 − 2 + 2 + 1 − 1 + + + +1 + 0 − 2 + 0 + 1 * *1 + 0 − 2 + 0 + 1*!0! m1 = 1 m 4 − 2m 2 + 1 = 0 (m 2 − 1)(m 2 − 1) = 0 m 2 = 1 ____ m 2 = 1 m 2 = 1 ___ m 2 = 1 m2 = +1 ____ m 4 = +1 m3 = −1 ____ m5 = −1 y = (C1 + C 2 x + C 3 x 2 )e x + (C 4 + C 5 x )e − x
19)
MECATRONICA
39
F.I.M.E.
MECATRONICA
20)
D 3 − 6 D 2 + 2 D + 36 y = 0 P q
±1± 2
=
+
−
− 2 ....1 − 6 + 2 + 36 ........... − 2 + 16 − 36 ..........1 − 8 + 18 + 0 m 1 = −2
(m
2
− 8m + 18
)
− b ± b 2 − 4ac 2a
=
− (− 8) ±
a + bi = 4 + 2i
(
(− 8) 2 2(1)
− 4(1)(18)
=
8 + 64 − 72 2
=
4± −8
= 4 ± 2 • −4i
)
y = C 1 e − 2 x + C 2 cos 2 x + C 3 sen 2 x e 4 x
dy − 7y = 0 dx D − 7y = 0 ( D − 7) y = 0 22)
m=7 y = c1e 7 x
d3y d2y dy 7 + + 12 =0 3 2 dx dx dx ( D 3 + 7 D 2 + 12 D ) y = 0 23. −
m 3 + 7 m 2 + 12m = 0 m( m 2 + 7 m + 12) = 0 m( m + 4)(m + 3) = 0 m1 = 0 m2 = − 4 m3 = −3 y = c1 + c2 e − 4 x + c3e −3 x
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MECATRONICA
24.-3(dy/dx)+11y=0 3dy`+11y=0 y(3m+11)=0 3m=-11 m=-11/3 y=C₁ℓ(-11/3)
EJERCICIOS 3.2
d 2 y dy + −6y = 0 dx 2 dx
1)
D2 + D − 6 y = 0 m2 + m − 6 = 0 ( m − 2)( m + 3) = 0 m=2 m = −3
y = C1 e − 3 x + C 2 e 2 x
2)
( D 2 + 2 D + 1) y = 0 → y = 1, Dy = −1, cuando : x = 0 ( D + 1) 2 y = 0 (m + 1) 2 = 0 m = −1 y = (c1 + c2 x )e − x Dy = −(c1 + c2 x )e − x + e − x c2 1 = (c1 + c2 (0))e − x 1 = c1 − 1 = −(c1 + c2 (0))e 0 + e 0 c2 − 1 = −c1 + c2 − 1 = −1 + c 2 c2 = 0 y = (1 + 0)e x y = ex
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F.I.M.E. 3) D²(D-1)y = 0
MECATRONICA y = 2, Dy = 3,
m² (m-1) = 0 m1 = 0 m2 = 1 y = c1e0x + c2ex y = c1 + c2ex Dy = c1 + c2e
D² = 2 ; x = 0
y = x + 2ex
28)
( D 3 − 4 D 2 + 4 D ) y = 0 ___ y = 1 m(m 2 − 4m + 4) = 0 ______ Dy = 2 ______________________ D 2 y = 8 ( m − 2) 2 = 0 (m − 2)(m − 2) = 0 m1 = 0 m2 = 2 m3 = 2 y = C1 + e 2 x (C 2 + C 3 x ) Dy = e 2 x * C 3 + 2(C 2 + C 3 x)e 2 x D 2 y = 2C 3 e 2 x + 4(C 2 + C 3 x)e 2 x + e 2 x * 2C 3 1 = C1 + C 2 _______________ C 3 = 2 − 2C 2 2 = C 3 + 2C 2 ________ 8 = 4(2 − 2C 2 ) + 4C 2 8 = 2C 3 + 4C 2 + 2C 3 __ 8 = 8 − 8C 2 + 4C 2 8 = 4C 3 + 4C 2 _______ 0 = −4C 2 2 = C3 + 0 C3 = 2 1 = C1 + 0 C1 = 1 y = 1 + 2 xe 2 x
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42
F.I.M.E. 4)
MECATRONICA
43
F.I.M.E.
MECATRONICA
5)
(D
2
)
+1 y = 0
m +1 = 0 m 2 = −1 m = −1 m = ±1i1 a + bi 0 +1 y = e ax [C1 senbx + C 2 cos bx ] y = e 0 x [C1 senbx + C 2 cos bx] y = C1 senx + C 2 cos x → solucion..general dy = C1 cos x − C 2 senx
− 1 = C1 cos(π ) − C 2 sen(π ) − 1 = C1 (−1) − 0 C1 = 1 1 =C 1 sen(π ) + C 2 cos(π ) 1 = C1 (0) + C 2 (−1) − 1 = C2 y = (1) senx + (−1) cos x y = senx − cos x
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6)
( D + 9) y = 0 _________________ y = 2 __ Dy = 0 ____________________________ cuando > x = π / 6 m2 + 9 = 0 m 2 = −9 m1 = 3i m2 = −3i a + bi Q + 3i y = e 0 x (C1 Sen3 x + C 2 Cos3 x) − − > solucion _ gral. y = C1 Sen3x + C 2 Cos3 x dy = 3C1Cos3 x − 3Sen3xC 2 2 = C1 Sen3(π / 6) + C 2 Cos3(π / 6) 2 = C1 + 0 C1 = 2 Dy = 3C1Cos3 x − 3Sen3 x − C 2 0 = 3C1Cos90 − 3Sen90C 2 0 = −3C 2 0 / 3 = C2 y = 2 Sen3 x + 0Cos3 x y = 2 Sen3 x
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F.I.M.E.
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7)
(D2 + 2D + 2)y = 0 ______y = 0 __Dy= 0 __cuando> x = 0 m2 + 2m + 2 = 0 − 2 + − 4 −8 2 m1 = −2 + (4)(−1) m1 = −1+1i m2 = −2 − − 4 2 (−2) − ( 4)( −1) − 2 − 2i = = −1− i 2 2 y = e−x (C1Senx+ C2Cosx 9
m2 =
Dy= e−x (C1Cosx− C2 Senx) + (C1Senx+ C2Cosx)e−x (−1) 0 = −C2 C2 = 0 1= C1 − 0 − 0 − C2 C1 =1 y = e−x (senx) EJERCICIOS 3.3
1)( D 2 − D − 2) y = e 3 x ( D − 2)( D + 1) y = 0 (m − 2)(m + 1) = 0 YC = c1e 2 x + c 2 e − x Yp = Ae 3 x Yp = ( D 2 − D − 2)( Ae 3 x ) = e 3 x D ( Ae 3 x ) = 3 Ae 3 x D 2 ( Ae 3 x ) = 9 Ae 3 x D 2 ( Ae 3 x ) − D ( Ae 3 x ) − 2( Ae 3 x ) = e 3 x 9 Ae 3 x − 3 Ae 3 x − 2 Ae 3 x = e 3 x 4 Ae 3 x = e 3 x 1 A= 4 YG = YC + Y p YG = c1e 2 x + c 2 e − x + 1 4 e 3 x 8 de julio de 2008
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F.I.M.E.
MECATRONICA
2)( D 2 − D − 2) y = sen(2 x ) ( m 2 − m − 2) (m − 2)(m + 1) = 0 YC = c1e 2 x + c 2 e − x YP = Asen2 x + Bsen2 x YP = ( D 2 − D − 2)( Asen2 x + B cos 2 x) = sen 2 x D ( Asen2 x + B cos 2 x) = 2 A cos 2 x − 2 Bsen2 x D 2 ( Asen2 x + B cos 2 x) = −4 Asen2 x − 4 B cos 2 x − 4 Asen2 x − 4 B cos 2 x − 2 A cos 2 x + 2 Bsen2 x − 2 Asen2 x − 2 B cos 2 x = sen 2 x − 6 Asen2 x − 6 B cos 2 x − 2 A cos 2 x + 2 Bsen2 x = sen2 x
− 6 A + 2B = 1 − 2 A − 6 B = 0(−3) 20 B = 1 1 B= 20
− 6 A + 2( A=−
YG = c1e 2 x + c 2 e − x −
1 ) =1 20
3 20
3 1 sen 2 x + cos 2 x 20 20
3)Y − $Y + Y − $Z = [\:
]; − 6] + 11] − 6 = 0 ] − 3] − 2] − 1 = 0 ]^ = 3 ] = 2 ]; = 1 ? = ^ 6 > + 6 > + ; 6 ;>
_ = `6 :; + a6 :> b; − 6b + 11b − 6`6 :> + a6 :> = 26 :> _c = `6 :> − `6 :> − a6 :> _cc = −`6 :> + `6 :> − `6 :> + a6 :> _ccc = `6 :> + `6 :> − `6 :> + `6 :> − a6 :> `6 :> + `6 :> − `6 :> + `6 :> − a6 :> + 6`6 :> − 6`6 :> + 6`6 :> − 6a6 :> + 11`6 :> − 11`6 :> − 11a6 :> − 6`6 :> − 6a6 :> = 26 :> 26`6 :> − 24`6 :> − 24a6 :> = 26 :> −24` = 2 1 `=− 12
26` − 24a = 0 26 − 24a = 0 − 12 13 a= 6−24
a=−
d = ^ 6 > + 6 > + ; 6 ;> −
13 144
1 13 :> 6 :> − 6 12 144
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47
F.I.M.E.
MECATRONICA
4) = 2 + − ] = 0 ∶ ]^ = 0: ] = 0
= ^ +
g = ` + a + h g = ` E + a ; + h b ` E + a ; + h = 9 + 2 − 1 b = 4` E + 3a + 2h = 9 + 2 − 1 b = 12` + 6a + h = 9 + 2 − 1 12` + 6a + h = 9 + 2 − 1 12 A=9 6B=2 2E=-1 A=9/12 B=2/6 E=-1/2 A=3/4 B=1/3 3 E 1 ; 1 ig = + − 4 3 2 3 1 1 ij = "^ + " + E + ; − 4 3 2 6) k − 1 = − lmn + + opq
]−5=0 ]=5 ? = ^ 6 r>
_ = ` − as6t + h + u cos _ = `s6t − as6t + h xs + u xs b − 5`s6t − as6t + h xs + u xs b = ` xs + `s6t − a xs − hs6t + h xs − us6t ` xs + `s6t − a xs − hs6t + h xs − us6t − 5`s6t + 5as6t − 5h xs − 5u xs = s6t − s6t + xs + xs
−h − 5`s6t = s6t ` − u + 5as6t = −s6t
−h − 5` = 1 −5 −5h+`=1 0h+26`=−4 −5h + ` = 1 : −5h + y z = 1 h=
:; ^;
` − 5h xs = xs −a + h − 5u xs = xs
`=
−2 13
^;
` − u + 5a = −1 h − 5u − { = 1
2 − u + 5a = −1 13 3 − − 5u − a = 1 13 −
−u + 5a = −
11 13
8 de julio de 2008
48
F.I.M.E. −5u − a =
MECATRONICA 16 13
.−u + 5a = − −5u − a =
16 13
11 / −5 13
71 0 − 26 a = 13 71 a=− 338 u = ` + 5a + 1
2 71 + + 5 *− ++1 13 338 69 u=− 338 u = *−
_ = ` − as6t + h + u cos i_ = .−
2 71 3 69 + / s6t − . + / xs 13 338 13 338
7) Y − 1Z = \ [ − [ +
_ = `6 > − a + h _c = `6 > − a `6 > − a − 5`6 > + 5a − 5h = 36 > − 2 + 1 −4` = 3 |=− 5a = −2 }=− 1 3 2 5 1− 3 −5 = ~ − −5h + a = 1 h = 1 −5 1 2 3 3 > d = ^ 6 r> + − − 6 5 25 4
]−5=0 ]=0 ? = ^ 6 r>
8) k − 1 = m − m1 ]=5 i = "^ 6 r> ig = 6 > − 6 r> ig = ` 6 > + a6 > + h6 > + u6 r> + 6 r> ig = ` 6 > + a6 > + h6 > + u 6 r> + 6 r> b` 6 > + a6 > + h6 > + u 6 r> + 6 r> = 6 > − 6 r> b` 6 > + 2`6 > + a6 > + a6 > + h6 > + 5u 6 r> + 2u6 r> + 56 r> + 6 r> = 6 > − 6 r> ` 6 > + 2`6 > + a6 > + a6 > + h6 > + 5u 6 r> + 2u6 r> + 56 r> + 6 r> − 5` 6 > − 5a6 > − 5h6 > − 5u 6 r> − 56 r> = 6 > − 6 r>
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F.I.M.E.
MECATRONICA
9)
(D − 1) y = sen( x) + cos(2 x) \ (D − 1) y = 0 (D − 1) = 0 m −1 = 0 m =1 Yc = C1e x Yp = Asenx + B cos x + E cos 2 x + Fsen2 x
(D − 1)( Asenx + B cos x + E cos 2 x + Fsen2 x ) = sen( x) + cos(2 x) D( Asenx + B cos x + E cos 2 x + Fsen2 x ) = D( Asenx + B cos x + E cos 2 x + Fsen2 x ) − 1( Asenx + B cos x + E cos 2 x + Fsen2 x ) = sen( x) + cos( 2 x) A cos x − Bsenx − 2 Esen 2 x + 2 F cos 2 x − Asenx − B cos x − E cos 2 x − Fsen 2 x = sen( x )+ cos( 2 x ) − Bsenx − Asenx = sen( x )................................................ − B − A = 1................ A = B A cos x − B cos x = 0......................................................... A − B = 0 2 F cos 2 x − E cos 2 x = cos( 2 x )......................................2 F − E = 1 − 2 Esen2 x − Fsen 2 x = 0............................................... − 2 E − F = 0 − B − B = 1.................................. − B − A = 1....................2 F − 1 = E 1 − 2 B = 1....................................... − 1 = A........................ − 2 E − F = 0 2 1 1 2 B = − ........................................ A = − ......................... − 2(2 F − 1) − F = 0.. → −5 F = −2..... → F = 2 2 5 − 2E − F = 0 2 2 1 − 2 E = ..... → ...E = − .... → ....E = − 5 10 5 2 1 1 1 Yg = − senx − cos x − cos 2 x + sen2 x 5 5 2 2
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F.I.M.E.
MECATRONICA
11) Y − Y + 1YZ = + 1 opq[
]; − 2] + 5] = 0 ]] − 2] + 5 = 0 = 1 { = −2 = 5 +2 ± −2 − 45 2 ± B−16 2 ± 4 = = = ±2 2 2 2 ? = ' ^ s6t2 + xs2) + ; _ = ` + `s6t2 + a xs2
_c = ` + `s6t2 + 2` xs2 + a xs2 − 2as6t2 _cc = 2` xs2 + 2` xs2 − 4`s6t2 − 2as6t2 − 2as6t2 − 4a xs2 _ccc = −2s6t2 − 2s6t2 − 4`s6t2 + 4` xs2 − 4a xs2 − 4a xs2 + 4a xs2 − 8as6t2
b; − 2b + 5b` + `s6t2 + a xs2 = 10 + 15 xs2 −2s6t2 − 2s6t2 − 4`s6t2 + 4` xs2 − 4a xs2 − 4a xs2 + 4a xs2 − 8ass6t2 − 4` xs2 − 4` xs2 + 8`s6t2 + 4as6t2 + 4as6t2 + 8a xs2 + 5` + 5`s6t2 + 10` xs2 + 5a xs2 − 10as6t2 = 10 + 15 xs2 −4s6t2 + `s6t2 + 14` xs2 + a xs2 − 18as6t2 − 8` xs2 + 8`s6t2 + 8as6t2 + 8a xs2 + 5` = 10 + 15 xs2 5` = 10
|=
14` + a − 8` + 8a = 15
6` + 9a = 15 12 + 9a = 15 } =
1 d = ^ s6t2 + xs2 + ; + 2 + 2s6t2 + xs2 3
8 de julio de 2008