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ANALISIS MATEMATICO II
TRABAJO PR.ACTICO N° 11 EJERCICIOS RESUELTOS
TEMA: Ecuaciones Diferenciales
V) Ecuacionf's diferend~lf's de primer orden
1) Metodo de variables separables:
p(x)dx - q(y)dy = 0 entonces Jp(x)dx - Jq(y)dy = c
a) dy = x2-I
dx y2-7 y2dy=(x2-I)dx -7 Jldy=J(x2-I)dx -7 lI3=x
3/3-x+c
b)y'=3.x2y
dy 2-7 - = 3.X Ydx1 2-7 -dy = 3.x dxy
-7 J..!.-dy= J 3.x2dxy
\ y=kex3
\
)1, 2 2_.2c -y =x +x y3
_1 dy =x 2(1 +l) -7 _1_dy=3.x2dx -7 J ~dy=J3.X2dx3 dx . 1+ y2 1+ Y
-7 arctgy = x3 + c
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d) 3.x.y dy + 4.y2dx= dx
~ 3xydy = (1 - 4.r)dx ~ y dy = _1 dx ~ I y dy = I_I dxI-4y2 3x 1-4y2 3x
1 2 1 . .2 - 8~ - - In(1 - 4.y ) = -lnx + c ~ In(1 - 4.y ) = -lnx + Ink833
~ In(1 - 4.y2)= In( k. x -8/3 ) ~ 1 - 4.y2= k. X ·8/3
j4.y2 = 1 _k. x -8731
e)2.y.y'+y'-4.x=3.x2'+2 si y(O)=-I
~ y'(2y+l)=3.x2+4x+2 ~ (2y+l)dy=(3.x2+4x+2)dx
-7 f(2y+l)dy= f(3.x2+4x+2)dx -7 y2+y=x3+2x2+2x+c -70=O+c
f) y' = 2. .JY+i .cos x-7 ~ dy = cosx.dx ~ f ~ dy = f cosx.dx -7 .JY+i = senx + c
2 y+l 2 y+I
~ Y= (senx + C)2 - 1 -7 0 = c2 - 1 -7 c = 1
g) dy - tgx.dx - y-.tgx.dx = 0 si y(O)=J31-7 dy = tgx.dx + y2.tgx.dx -7 dy = (1 + r)tgxdx -7 --dy = tgxdx
1+ y2
III III senx. -7 -- dy = tgxdx -7 -- dy = -- dx ~ arctgy = -lnlcosxl + c1+ y2 1+ y2 cosx
-7 arctg J3= -lnlcosOI+ c ~ n/3 = 0 + C
I y = tg[rr/3 - In(cosx)] I
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a) x2 - .; = e es una familia de hiperbolas
~ 2x-2yy' =0 ~ y'= x ~ dy = _ x..y dx x
...>,( ..j ·.:f····1. .,,\ ::'~"'" )<'.,(V'" ··f ,· . '.J»~'?" ~;t ~ ~ t~./.,.. ;I.\. )( \ ?t.. .~.. l.' ,,' I l.~ \ \ / '1" ..... ,!.'t \ ~~ . ~ / •.. "-+-~..... l: t ~ _
_, :1 ~.I1 :,: .;~;'! ,: f'.... ~ t' \\\\-?>x<..Y··· X<·:~,?<::~\,:>.:: ::..>:::..:~>~.:~.:: ..~...~. ··.,..·..£I·.>.f.j
1 1-7 - dy = - - dx ~ Iny = - luxy x
-7 Iny = Inx·' + k = Inx -I + 1m = In(rx -I)
-7 y = rx -I -7 y = r Ix familia de hiperbolas
b) y2 = ex familia de parabolas2
~ 2yy' = e ~ 2yy' = L ~ y,=Lx 2x ~~.~~
" \.2 \. -1 :;Jf. 1 / 2 /
~~ •.~
~ dy = _ 2xdx Y
Y2~ _ =_x2+k2
familia de elipses
2 2 3 dy _ 2xe) y2=cx3 ~ 2yy' =3cx2 =3L x2=3L ~ y' =...1.. ~ ----x3 X 2x dx 3y
2 2~ 3ydy = -2xdx ~ 3L = - x2 + k ~ 3L + x2 = k elipses
2 2
y dy _ xd) xy = c ~ l.y + xy' = 0 ~ y' = - - ~ - - - ~ ydy = xdx
x dx yy2 X 2 2 2
~ - = - + k ~ L -~ = k familia de hiperbolas2 2 2 2
2 2 ~ , ~ x _~ dy = 2ye) x + 2y = c -, 2x + 4yy = 0 -, y' = -- -,2y dx x
112-7 -dy = 2-dx -7 Iny = 21nx + k = 21nx + 1m = In(rx ) -7y x
y = rx 2 familia de parabolas
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P(x,y)dx + Q(x,y)dy = 0 tal que 8P = 8Q continuas, entonces existe f(x,y) tal queOy Ox
df= P(x,y)dx + Q(x,y)dy entonces f(x,y) = c (1)
8P 8Q_. = ~cosx.seny = -Oy Ox
f(x,y) = fp(x,y)dx = f (cosx.cosy + 2x)dx = senx .cosy + x2 + g(y)
~ f Y(x,y) = -senx.seny + g'(y) = Q(x,y) = - (senx.seny + 2y) ~ g'(y) = -2y
~ g(y) = - y2 + k ~ f(x,y) = senx .cosy + x2-l + k ~ por (1)
senx .cosy + x2-l + k = c
I senx .cosy + x2 -l = c* I
b) (eXy+ x eXy)dx + (x eX+ 2)dy = 0 si y(O) =-1
8P 8Q_ =eX + x.ex =_Oy Ox
f(x,y) = fceXy+ x eXy)dx = eXy+ xeXy - eXy+ g(y) = x eXy+g(y)
f y(x,y) = x eX+ g'(x) = Q(x,y) = x eX+ 2 -) g(y) = 2y + k -) x eXy+ 2y = c*
si x = 0, y = -1~ c* = -2
c) cosedr + (eo __r.sene)de = 0
Q(e, r) =eo-r.sene 8P = -sene = 8Q8S ar
g(S) ·~f a(S , r) = - senS.r + g'(S) = Q(S , r) = eO- r.senSf(S , r) = f cosSdr = r.cosS
-7 gee) = eO + k·1[(ft, r)= r.cosS+ eH= c* I
d) [2x + y2 - cos(x + y)]dx + (2xy - cos(x + y)- eY]dy= 0
P(x,y) = 2x + l- cos(x+ y) Q(x,y) = 2xy - cos(x + y)- eY
8P . 8Q- = 2y + sen(x + y) = -Oy Ox
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[(x,y) = J[2x -I- y2 - COS(X + y)]dx = Xl} Xy2-- scn(x -I- y) + g(y) ~
r y (X,y) = 2xy - COS(X -I- y) -I· g'(y) = Q(X,y) = 2xy - COS(X \- y)- eY ~ g(y):::: -eY
a) dy _ y ::::e3x ~ y' _ y = e3xdx
u'v -I- uv' -uv = e3x ~ v(u'- u) -I- uv' = e3x (1)
-7. si u' -- u = 0 ,
~ v = e2x 12 +c
en (1) queda eX .v' = e3x -) dv = e3x e -l( dx = e2x dx...•• x 2x /"7 porser y=u.v=c' (e' 2+c)
y ~ e~' + 0'.0 Ib) y' -I- tgx. Y = secx
~ y=u.v ~ u'v-l-uv'-; tgx.u.v=secx -7 v(u'+tgx.u)+uv'=secx (1)
Si u' -I- tgx .u= 0 -7 u' = - tgx.u -) J.. du = -tgxdx ~ Inu = In(cosx) ~ 11 = COSX ]11
._----~--------------
. '-I- ( 0 (1) d 'e x -"- (Iv = sec2x dxSIll gx.u= ,en que a cosx.v=s-c. "7 .
~ Y = u.V = cosx (tgx -I- c)
EsenX-l-c.c~
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c) x dy + 2y = 5.x3dx
x.y' + 2y = 5.x3 ~ y'+ 2y = 5.x2 ~ con y =uv (*) y'= u'v + uv;X
2u'v + uv'+ -uv = 5.x2
X
2v(u' + -u) + uv'= 5.x2
X
2u'+ -u=O
x2 1 2
~ u'=- -u ~ -du=- - ~ Inu=-2Inx=lnx02 ~ u=x02x U X
(1) queda x 02v'= 5.x2
y'_ 5y = _1 xy3 (*)2
z = y'-n ~ Z' = (1- n)y·n .y' ~ como en este caso n =3 .~ z == y.2 (**)
Z' = -2y -3.y' ~ en (*) multiplicamos por -2.y -3 para poderexpresar en terminos de
esta ecuaci6n diferenciallineal tomando z = uv z' = u'v + uv' asf se obtiene que
u = e-'ox v = x elOx/2- e,ox/20 + c z = e-'OX(xe,ox/2 - e,ox/20 + c)
por (**) y -2= Z = e-10\x elOx/2- e,ox/20 + c)
)' y2 + 2xy
e y = x2, ~2 2 2 0y -x y --y=
x
, 2. -2 2Y --y =x.> yx
(1) tomo z == yl-n con n == 2 (Bernoulli)
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multiplico (1) por y-2
, 2 -2-z --z=x -+x
u'v + uv' + 3.uv = - x-2X
queda uv'= - x-2 -+
, 2 -2Z +-z=-x
x
-+ V ( u' + ~u J + uv' = - x-2
x-2v' = - x-2 -+ dv = -dx
, 2u +-u=O
X
entonces z==X-2(_ X + c) ~
I y.[ =x·z(_ x + c) I
f) dy .·-2-.-+ 2y=xydxz'= 3y2y' multiplicando en(l) por 3y2 queda la ecuaci6n diferenciallineal
3y2y' + 2y3y2 = xy-23y2 -+ z' + 6z = 3x -+ tomando z = uv
u'v+uv'+6uv=3x -+ v(u'+6u)+uv'=3x -+ si u'+6u=O -+ u=e-6x
e6x e6x xe6x e6x
uv'=3x -+ e-6Xdv=3xdx -+ v= J3xe6Xdx=3x7-3 36 +c=2-12+c
y3 = x __1 + ce-6x2 12
Dada la ecuac.i6n diferencial P(x,y)dx + Q(x,y)dy =0 sent hotl1ogeneasi cumple que:
P(tx,ty) =t"P(x,y) , Q(tx,ty) = t ".Q(x,y) es decir, sf P y Q son funciones homo gene as
de igual grado.
Para resolver se realiza la sustituci6n y = ux 0 x = uy segun el casocon el objetivo de
transformarla en una ecuaci6n diferencial de variables separables.
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'a) y2_ x.y + x2.y' =0 -7 (y2 - x.y)dx + x2.dy = 0 (I)
7 P(tx,ty) = (ty)2 - tx.ty = t2 (y2 - x.y) = t2P(x,y); Q(tx,ty) = (tx)2 ==t2 x2 = t2.Q(x,y)
n ==2 si y = ux queda dy = du.x + u.dx sustituyendo en (I) :
[ (UX)2 _ x.(ux)]dx + x2.( du.x + u.dx ) = 0 -7 ux2 (u - I)dx + x3du + u. x2dx ==0
-7 ux2(u-1 + l)dx+x3du=0 -7 u2x2dx =-x3du -7 !dx=- _I_dux . u2
17 Inlxl + c = - ~u
1 1u = --- ~ y = "X
In Ix I+c In Ix I+c
b) (xy + x2 + y2)dx = x2dyP(tx,ty) = ep(x,y) ; Q(tx,ty) = t2Q(x,y) -7 n = 27 y ==ux dy= du.x + u.dx
7 (xux + x2 + (uxi)dx = x2 (du.x + u.dx) -7 x2 (u + u2 + l)dx = x3 du + u. x2dx
-7 x2(U+U2+ l-u)dx=x3du -7 !dx= __ l_du -7 Inlxl+c=arctgux 1+ u2
dy x2 _y2c)-=--
dx 3.x.y
comprobar que es homogenea con n = 2
reemplazando en Ia ecuaci6n y =ux dy = du.x + u.dx se obtiene:
3xux(du.x + u.dx) = (x2 - (ux)2)dx -7 3ux3du = x2dx (1 - u2_ 3u2)
3u 1 37 --du = - dx -7 -=-In(l -4 u2) = lnx + c = In (kx) 7 (l ~4u2y3/8 = kx1- 4u3 X 8
-7 1 -4 u2= (kxy8/3 -7 como y = ux u = y/x -7 1 -4(y/xi = (kxr8/3
-7 (x2 _ 4y2).X-2= k'.x -8/3 -7 x2 - 4y2= k'.x -2/3 -7 (x2 - 4Y). X 2/3 == k'
, y2 +X~X2 +y2d) y = ------ comprobar que es homogenea y su soIuci6n es :
3x.y
)-m' ~ln~l+c
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1.
a) y" - 5y' + 6y = 0_,,- 2x 3x-, Y = CI e + C2 e
b) y"- 5y' = 0_:'l. + 5x-, Y = C, C2 e
c) y"-2y'+5y=0 ~ 'r2-2r+5=0 ~ r,=1+2i
~ y = eX( Acos2x +Bsen2x)
e) y"+ 4y' + 4y = 0 ~ r2 + 4r + 4 = 0 ~ r, =-2
~ Y == (CI + C2 x) e -2x
f) y" - y' - 6 y = 0 ~ r2 - r- 6 = 0 -7 r, = -2 r2 = 3_:'l. -2x + 3x-, y= Cj e C2 e
2. Resolver utilizando el metodo de los coeficientes indeterminados: ay" + by' + y = f(x)
a)y"-y=-llx+l ~ r2-1=0 -7 rl=1 fO
( porque f(x) es un polinomio )x - xYg = CI e + C2 e
u=Ax+B ~u'=A -7 u"=O -7-Ax-B=-llx+17 A=ll B=-1
x -2"yg = c, e -I-- C2 e
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u= Ax2+Bx+C -7 u'=2Ax+B -7 u"=2A
-7 2A + (2Ax + B) -2(Ax2 + Bx + C) = x2-2x + 3
{
-2A= 1 A=-I/2
2A - 2B = -2 B = 112
2A + B - 2C = 3 C = -7/4
)" '12 2x ~ 2 12 0 ~ 4 ~ 2c Y - Y - Y= e "7 r - r- = "7 r) = -r~ ~
exponencial )~ 4x -3x7' Yg = CI e + C2 eu=Ae2x -7 u'=2Ae2x -7 u"=4Ae2x
-7 4Ae2x _2Ae2x -12Ae2x =e2x -7 A=-1II0
Y= CIe 4x+ C2e-3x - (1/10) e 2x
d)YU_Y'_12y=e4~ r2-r-12=0 -7 r) =4 jojo! r2=-3 1'44x - 3xYg= CIe + C2e
u = A e 4x.x (aparece x como factor)
-7 u'=Ae4x +4Ae4x.x = Ae4x (1 +4x) -7 u"=4Ae4x (l +4x)+4Ae4x
-7 4 A e 4x (l + 4x) + 4 A e 4x - A e 4x (I + 4x) -12 A e 4x.x = e 4x
7 Ae4x=e4x -7 A= 1/7
Y= c, e 4x+ C2e - 3x + (l/7).x e 4x
x -3xYg= Ct e + C2e
e) y" + 2y' - 3y = 7cos(3x) -7 r2 + 2 r - 3 = 0 -7 rl = 1 ;t 3i~ ~
u = Asen3x + Bcos3x -7 u' = 3Acos3x -3Bsen3x -7 u" = -9Asen3x - 9Bcos3x
-9Asen3x - 9Bcos3x + 2(3Acos3x -3Bsen3x ) ~ 3(Asen3x + Bcos3x) =7cos(3x)
{
. -12A- 6B ~O B ~-2A
6A-12B=7 6A+24A=7 A=7/30 B=-7/15
Y= CI e x+ C2e - 3x +(7/30)sen3x - (7/15)cos3x
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f) y" - 5y' + 6y = x.ex ~ r2 -5 r + 6 = 0 ~ r, = 3 t 13x 2xYg = c, e + C2 e
u = (Ax + B) eX ~ u' = A eX + (Ax + B) eX= eX(A + B +Ax)
~ u"= eX(A+B+Ax)+ex A =eX(2A+B+Ax)
~ eX(2A + B +Ax) - 5 eX(A + B +Ax) +6 eX(B +Ax) ::::x eX
~ eX(2A- 5A + B - 5B + 6B + Ax -5Ax + 6Ax) = x.ex
~{. 2A= 1 A= 1/2
. . •-3A + 2B = 0 B = %
g)y"_4y'+4y::::x.e2X ~ r2-4r+4=O ~ r,:::: r2=2 ojo!!
yg = (c, + C2x)e2X
u = x2(Ax + B) e 2x = e2x(Ax3 + Bx2) (se tuvo que multiplicar por x2 )
~ u':::: 2e2x(Ax3 + Bx2) + e2x(3Ax2 + 2Bx) ::::e2x(2Ax3 + 2Bx2 + 3Ax2 + 2Bx)
~ u":::: 2 e2x(2Ax3 +2Bx2 + 3Ax2 + 2Bx) + e2x(6Ax2 + 4Bx + 6Ax + 2B)::::
::::e2X(4Ax3+ 4Bx2 + 12Ax2+ 8Bx + 6Ax+ 2B)
e2x[4Ax3+ 4Bx2 +12Ax2+ 8Bx+6Ax+2B - 4(2Ax3+2Bx2+3Ax2+2Bx)+ 4(Ax3+Bx2)]:::: x.e2x
e2x(4Ax3 + 4Bx2 +12Ax2+ 8Bx+6Ax + 2B - 8Ax3-8B~-12Ax2- 8Bx + 4Ax3+ 4Bx2) = x.e2x
~ e2X(6Ax + 2B):::: x.e2x ~ A = 1/6 B = 0
u::::X (Asenx +Bcosx) ~ u' = Asenx + Bcosx + x(Acosx -Bsenx)
~ u":::: Acosx - Bsenx +Acosx - Bsenx + x(-Asenx -Bcosx) ::::
::::2Acosx - 2Bsenx + x(-Asenx -Bcosx)
2Acosx - 2Bsenx + x(-Asenx-Bcosx) + x (Asenx + Bcosx) = senx