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course Mth 279
2/20 9
Section 2.2*********************************************
Question: 1. Solve the following equations with the given initial conditions:
1. y ' - 2 y = 0, y(1) - 3
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Your solution:
this y(1) - 3 I think should be an = sign.
I'm going to let y(1) be y(x), and sub y' = dy(x) / dx
dy(x)/dx = 2y(x)
2 = dy(x)/dx/y(x)
now integrate each side wrt x
2x + c = log(y(x))
simplify; y(x) = e^(2x + c)
y(x) = ce^(2x) now substitute y(x) = 3
3 = ce^2
solve for c = 3/e^2 which is roughly 0.41
putting c back into the y(x) equation
y(x) = (3/e^2)(e^2x) = 3e^x
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y = 3 e^x, when substituted into the given equation, gives us
y ' - 2 y = 4 e^x,
not
y ' - 2 y = 0.
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Question: 2. t^2 y ' - 9 y = 0, y(1) = 2.
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Your solution:
substitute y' = dy(t)/dt and separate the equation to where we can integrate it simply enough.
t^2(dy(t)/dt) = 9y(t)
9/t^2 dt = dy(t)/dt / y(t) dt
integrate this wrt t.
-9/t + c = log(y(t))
y(t) = e^(-9/t) + c
now sub y(t) = 2 in to solve for the constant
2 = c* e^-9
c = 2/e^-9 simplified. c = 2e^9 roughly 16,206.16
now you could put that back into the y(t) equation.
y(t) = 2e^(-9/t +9)
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Given Solution:
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Good.
You should also be able to solve this as a linear homogeneous equation of form y ' + p(t) y = 0.
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Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.
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Your solution:
sub y' as dy(t)/dt and fill in, separate to integrate the easiest, there is a lot of typing so i'll skip to my integrand
int( dy(t) / dt / y(t) ) dt = - int( (2t+1) / (t^2 + t) ) dt ------- both sides have the form of 1/u du = log u
becomes log(y(t)) = - log(t) - log(t+1)
y(t) = -e^t - e^(t+1+c) --------- this might just combine 1 and c to just c, anyways,
y(0) = -e^0 - e^(1 + c)
1 = -1 - ce^1
c = 2e^1
fill into y(t) equation to get ( i think)
y(t) = -e^t - 2e^(t+1)
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Given Solution:
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Your approach would work but you haven't integrated (2 t + 1) / (t^2 + t) correctly. The result is ln | t^2 + 1 |.
However you also need to solve this using the technique of linear homogeneous equations, which in fact requires the same integral with a slightly different concept.
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Question: 4. y ' + sin(3 t) y = 0, y(0) = 2.
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Your solution:
sub y' to dy(t)/dt and fill in, separate to integrate easily, lots of typing, skip to integrand.
int ( dy(t)/dt/y(t) ) dt = - int ( sin(3t) ) dt ----- both again have the form int 1/u du = logu
becomes
log(y(t)) = 1/3cos(3t) + c
y(t) = 1/3e^(cos(3t) + c)---------- I think, my 1/3 might be incorrect
y(0) = 1/3ce(cos(0))
2 = 1/3ce^1 c = 6/e^1
fill back into the y(t) equation
if my math is correct in multiplying my constant with e that is.........
y(t) = 2/e^1 * e^(cos(3t))
???????????????? I'm getting the hang of this, just not completely sure of my multiplication near the last couple steps.
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Given Solution:
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Your constant isn't right, because e^0 = 1, not e. Fix that and your solution will be OK.
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Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.
y ' - t^2 y = 0
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y ' - y = 0
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I'm choosing graph A for this, because the derivative is = to y, which is the rise over the run of y. This graph is the only one that represents an exponential change.
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y' - y / t = 0
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This graph is similar to a y=x graph. when the change in y wrt t is equal to t on the right, it will equal itself. so this graph should resemble a straight line, which is why i'm choosing graph C
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y ' - t y = 0
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graph F, because this is an opposite wave pattern than graph B, the sign is just reversed, not sure about this.
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y ' + t y = 0
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I'm choosing graph B for this, because it changes much more dramatically than graph A. It looks to be asymptotic.
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A
B
C
D
E
F
6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?
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Your solution:
the slope of this graph is 3. change in y wrt change in x
by doing like before, and sub y' with dy(x)/dx and separating to integrate.
-bx + c = log y(x)
y(x) = e^(-b + c)
y'(x) = -be^(-bx+c)
now sub 3 in
3 = -bce^(-b)
b = -3ce^b
?????????? i'm not absolutely sure about this, because i'm getting that b is a function of b, but I think its not a shabby attempt.
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Given Solution:
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Question:
7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.
If we let w(t) = y(t) + 2, then:
What is w ' ?
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sub y' as y(t)/dt and rearrange to find that w' = y' = dy(t)/dt
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What is y(t) in terms of w(t)?
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since dy(t)/dt = w(t), derive w(t) to get,
dw(t)/dt = dy(t)/dt /dt
dw(t) = y(t)dt
w'(t) = y(t)dt
y(t) = dw(t)/dt / dt
i think
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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?
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w(t) - dw(t)/dt /dt = 2
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Now solve the equation and check your solution:
Solve this new equation in terms of w.
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w(t) = 2+ dw(t)/dt/dt
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Substitute y + 2 for w and get the solution in terms of y.
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y+2 = 2+dw(t)/dt/dt
y = dw(t)/dt/dt
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Check to be sure this function is indeed a solution to the equation.
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I'm not sure about the /dt/dt part. maybe an error somewhere, assistance?
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Your solution:
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Given Solution:
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Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?
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Your solution:
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Given Solution:
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end document
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I've got a computer malfunction right now and can't send you the link I want to send. Next document you submit ask about the response to query_01.
Your work isn't bad but you need to be able to use the techniques of linear homogeneous equations. Gotta build on that technique. Check my notes.
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