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Catégorie :Category: mViewer GX Creator Ndless TI-Nspire
Auteur Author: hillpeter
Type : Document nécessitant un lecteur
Page(s) : 22
Taille Size: 563.94 Ko KB
Mis en ligne Uploaded: 03/05/2017 - 17:34:52
Uploadeur Uploader: hillpeter (Profil)
Téléchargements Downloads: 31
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a938509
Type : Document nécessitant un lecteur
Page(s) : 22
Taille Size: 563.94 Ko KB
Mis en ligne Uploaded: 03/05/2017 - 17:34:52
Uploadeur Uploader: hillpeter (Profil)
Téléchargements Downloads: 31
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a938509
Description
MATH 267 Final Study Guide S ’17
1 First Order Equations
dx π
Q1. = 4(x2 + 1); x = 1.
dt 4
Solution. This is a separable equation, so we proceed using the method of separation of variables.
dx
= 4(x2 + 1)
dt
dx
2+1
= 4dt
Z x Z
dx
= 4dt
x2 + 1
arctan x = 4t + C.
Here we utilize our initial condition to determine the value of C.
C = arctan x − 4t
π
= arctan (1) − 4
4
π
= −π
4
−3π
= .
4
We now finish our derivation of an explicit solution.
3π
arctan x = 4t −
4
3π
x(t) = tan 4t −
4
♦
1
MATH 267 Final Study Guide S ’17
dy π
Q2. − (sin x)y = 2 sin x; y = 1.
dx 2
Solution. This is a linear equation, so we proceed by finding an integrating factor. Note this equation is already in standard
form. Let µ denote the integrating factor. Then
R
P (x)dx
µ=e
R
− sin xdx
=e
= ecos x .
Multiplying our differential equation by the integrating factor, we can solve for a general solution.
dy
ecos x − sin x ecos x y = 2 sin x ecos x
dx
(ecos x y)0 = 2 sin x ecos x
Z
cos x
e y = 2 sin x ecos x dx
ecos x y = −2 ecos x +C
y = −2 + C e− cos x .
Finally, we make use of our initial condition to determine the value of C and derive a solution to the IVP.
1 = −2 + C e− cos (π/2)
3 = C e0
3 = C.
y(x) = −2 + 3 e− cos x
♦
2
MATH 267 Final Study Guide S ’17
Q3. (x + y)2 dx + (2xy + x2 − 1)dy = 0; y(1) = 1.
Solution. This equation appears to be exact, so we check to be sure.
(x + y)2 dx + (2xy + x2 − 1)dy = 0
(x2 + 2xy + y 2 )dx + (2xy + x2 − 1)dy = 0
∂ 2
My = (x + 2xy + y 2 ) = 2x + 2y
∂y
∂
Nx = (2xy + x2 − 1) = 2y + 2x
∂x
M y = Nx .
Therefore, this is an exact equation so we proceed using that method.
Z
F (x, y) = M dx
Z
= (x2 + 2xy + y 2 )dx
1 3
= x + x2 y + xy 2 + h(y).
3
Using N , we solve for h(y).
N = 2xy + x2 − 1
∂
N= F (x, y) = x2 + 2xy + h0 (y)
∂y
2xy + x2 − 1 = 2xy + x2 + h0 (y)
h0 (y) = −1
h(y) = −y + C.
Putting these together, we have a general solution to the equation.
1 3
x + x2 y + xy 2 − y = C.
3
Using our initial condition, we solve for the value of C and derive a solution to the IVP.
1 3 4
C= (1) + (1)2 (1) + (1)(1)2 − 1 =
3 3
1 3 4
x + x2 y + xy 2 − y =
3 3
♦
3
MATH 267 Final Study Guide S ’17
Q4. (2y 2 + 3x)dx + 2xydy = 0.
Solution. This equation appears to be exact, so we check to be sure.
∂
My = (2y 2 + 3x) = 4y
∂y
∂
Nx = (2xy) = 2y
∂x
My 6= Nx .
Therefore, this is not an exact equation. We check for a possible integrating factor to make it exact.
M y − Nx 2y 1
= = .
N 2xy x
R
(1/x)dx
µ=e = eln x = x.
Multiplying the equation by µ should make it exact.
(2xy 2 + 3x2 )dx + 2x2 ydy = 0
∂
My = (2xy 2 + 3x2 ) = 4xy
∂y
∂
Nx = (2x2 y) = 4xy
∂x
M y = Nx .
The modified equation is exact, so we proceed using that method.
Z
F (x, y) = N dy
Z
...
1 First Order Equations
dx π
Q1. = 4(x2 + 1); x = 1.
dt 4
Solution. This is a separable equation, so we proceed using the method of separation of variables.
dx
= 4(x2 + 1)
dt
dx
2+1
= 4dt
Z x Z
dx
= 4dt
x2 + 1
arctan x = 4t + C.
Here we utilize our initial condition to determine the value of C.
C = arctan x − 4t
π
= arctan (1) − 4
4
π
= −π
4
−3π
= .
4
We now finish our derivation of an explicit solution.
3π
arctan x = 4t −
4
3π
x(t) = tan 4t −
4
♦
1
MATH 267 Final Study Guide S ’17
dy π
Q2. − (sin x)y = 2 sin x; y = 1.
dx 2
Solution. This is a linear equation, so we proceed by finding an integrating factor. Note this equation is already in standard
form. Let µ denote the integrating factor. Then
R
P (x)dx
µ=e
R
− sin xdx
=e
= ecos x .
Multiplying our differential equation by the integrating factor, we can solve for a general solution.
dy
ecos x − sin x ecos x y = 2 sin x ecos x
dx
(ecos x y)0 = 2 sin x ecos x
Z
cos x
e y = 2 sin x ecos x dx
ecos x y = −2 ecos x +C
y = −2 + C e− cos x .
Finally, we make use of our initial condition to determine the value of C and derive a solution to the IVP.
1 = −2 + C e− cos (π/2)
3 = C e0
3 = C.
y(x) = −2 + 3 e− cos x
♦
2
MATH 267 Final Study Guide S ’17
Q3. (x + y)2 dx + (2xy + x2 − 1)dy = 0; y(1) = 1.
Solution. This equation appears to be exact, so we check to be sure.
(x + y)2 dx + (2xy + x2 − 1)dy = 0
(x2 + 2xy + y 2 )dx + (2xy + x2 − 1)dy = 0
∂ 2
My = (x + 2xy + y 2 ) = 2x + 2y
∂y
∂
Nx = (2xy + x2 − 1) = 2y + 2x
∂x
M y = Nx .
Therefore, this is an exact equation so we proceed using that method.
Z
F (x, y) = M dx
Z
= (x2 + 2xy + y 2 )dx
1 3
= x + x2 y + xy 2 + h(y).
3
Using N , we solve for h(y).
N = 2xy + x2 − 1
∂
N= F (x, y) = x2 + 2xy + h0 (y)
∂y
2xy + x2 − 1 = 2xy + x2 + h0 (y)
h0 (y) = −1
h(y) = −y + C.
Putting these together, we have a general solution to the equation.
1 3
x + x2 y + xy 2 − y = C.
3
Using our initial condition, we solve for the value of C and derive a solution to the IVP.
1 3 4
C= (1) + (1)2 (1) + (1)(1)2 − 1 =
3 3
1 3 4
x + x2 y + xy 2 − y =
3 3
♦
3
MATH 267 Final Study Guide S ’17
Q4. (2y 2 + 3x)dx + 2xydy = 0.
Solution. This equation appears to be exact, so we check to be sure.
∂
My = (2y 2 + 3x) = 4y
∂y
∂
Nx = (2xy) = 2y
∂x
My 6= Nx .
Therefore, this is not an exact equation. We check for a possible integrating factor to make it exact.
M y − Nx 2y 1
= = .
N 2xy x
R
(1/x)dx
µ=e = eln x = x.
Multiplying the equation by µ should make it exact.
(2xy 2 + 3x2 )dx + 2x2 ydy = 0
∂
My = (2xy 2 + 3x2 ) = 4xy
∂y
∂
Nx = (2x2 y) = 4xy
∂x
M y = Nx .
The modified equation is exact, so we proceed using that method.
Z
F (x, y) = N dy
Z
...
