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course MTH 279
1. y ' - 2 y = 0, y(1) - 3y = e ^ (-2t)
y’ = -2e^(-2t)
-2Ce^(-2t) + 2C e ^ (-2t) = 0
y(1) - 3 = -2Ce^(-2) + 2C e ^ (-2) = 3
C = 3
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Your solution doesn't work in the equation. It would be easy to make a slight adjustment in your solution method and get it to work.
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y = C e^(2t) is the general solution, as you can verify.
y(1) = 3 if C e^2 = 3, so that C = 3 / e^2.
So
y = 3 / e^2 * e^(2 t) = 3 e^(2t - 2).
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2. t^2 y ' - 9 y = 0, y(1) = 2.
y ' - 9 / t^2 * y = 0
y = e^(-9/t)
y’ = 9/t^2 * e^(-9/t)
9/t^2 * Ce^(-9/t) - 9/t^2 * Ce^(-9/t) = 0
y(1) = 9/t^2 * Ce^(-9/t) - 9/t^2 * Ce^(-9/t) = 2
C = 2
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Good. You have effectively shown that y = C e^(-9/t) is a solution to the equation.
However the C = 2 solution does not satisfy the initial condition. With C = 2 we get y(1) = 2 e^(-9), which is not equal to 2.
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9/t^2 * 2e^(-9/t) - 9/t^2 * 2e^(-9/t) =
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3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.
y' + (2t + 1)/ (t^2 + t) * y = 0
u = (t^2 + t)
du = (2t + 1)
y = e^(-ln(t^2+t)
y = - (t^2 + t)
y’ = (2t + 1) * (t^2 +t)
(2t + 1) * (t^2 +t) - (t^2 + t) (2t + 1) = 0
C = 1
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Good, but C = 1 does not give you a solution that satisfies the initial condition y(0) = 1.
In fact, y(0) is not even defined.
The initial condition y(0) = 1 cannot be applied, since the equation is not defined for t = 0 (nor is the general solution).
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4. y ' + sin(3 t) y = 0, y(0) = 2.
y = e^ (1/3 cos (3t))
y’ = -sin(3t)e^(1/3 cos(3t))
-sin(3t)e^(1/3 cos(3t)) + Ce^ (1/3 cos (3t)) = 0
y(0 = 2
-sin(3(0))e^(1/3 cos(3(0))) + Ce^ (1/3 cos (3(0))) = 2
C = 2/ e^1/3
-sin(3t)e^(1/3 cos(3t)) + 2/e^1/3 * (e^ (1/3 cos (3t))) = 0
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Your solution could be written
C e^(cos(3 t) / 3).
y(0) = 2 yields
C e^(cos(3 * 0) / 3) = 2
so that C = 2 e^(-1/3).
Your expression
2/e^1/3 * (e^ (1/3 cos (3t)))
would be the general solution.
You haven't clearly indicated that this is the general solution.
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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
E, the only direction field where the slope changes to zero and then back to what it began as, this is due to the t^2 not being affected by negative numbers so it makes sense the left and right hand side will be the same
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I believe that this equation corresponds to graph B.
y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative
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y ' - y = 0
A, the only direction field where the slope stays the same across the t axis, this is due to the equation not having a t in it
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I agree.
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y' - y / t = 0
C, because y' = y / t, the slopes will be getting smaller and smaller due to t growing faster than y
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y ' - t y = 0
B, because when finding the slope, y' = t y, in the first quadrant the slopes will be positive and the fourth quadrant negative
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I believe this one matches graph D.
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y ' + t y = 0
F, because y' = -t y, in the first quadrant the slopes will be negative and the fourth quadrant negative
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y ' = y / t; along any vertical line slope is equal to y, so magnitude of the slope increases as you move away from the x axis along such a line.
The slope increases in magnitude as you approach the y axis.
The slope is positive in the first and third quadrants, negative in the second and fourth.
Only graph C has these characteristics.
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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?
Ae^ -bt
y’ = -b e ^-bt
(8-2) / (3-1) = 3 = y' plug this into the original
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This is an average value of y ' over the interval from t = 1 to t = 3.
It cannot be applied at a specific point.
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We can use the two given points and the general solution to evaluate the two parameters A and b, by finding two simultaneous equations in A and b.
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The equation y ' + b y = 0 is first-order linear homogeneous, with solution y = A e^(-b t).
The graph passes through (1, 2), so
2 = A e^(- b * 1)
The graph passes through (3, 8), so
8 = A e^(-b * 3).
Thus we have two simultaneous equations in C and b.
Dividing the second by the first we obtain
e^(-2 b) = 4
so that
b = -1/2 ln (4) = ln (1/2).
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We can continue by finding A and expressing the corresponding specific solution.
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3 = e^-b(8) We will use the point (3,8) to solve for b
3 = e^(8b)
b = 1/8 ln(3)
using (1,2)
1 = Ae^-b(2)
1 = Ae^(1/4 ln (3))
-ln(3) = A
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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 ' ?
w' = (2+y) + 2 We get 2 + y from y'
= 4 + y
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What is y(t) in terms of w(t)?
y(t) = w(t) - 2
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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?
w’ - w +2 = 2
w’ - w = 0
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Now solve the equation and check your solution:
Solve this new equation in terms of w.
Here I would solve for a general solution either using the integrating factors or the general solution equation for non-homogeneous equations.
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Your solution is easily seen to be w = A e^t.
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Substitute y + 2 for w and get the solution in terms of y.
y’ - y = 2
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This is the differential equation.
You need to substitute y+2 for the w function in the solution w = A e^t.
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Check to be sure this function is indeed a solution to the equation.
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If you do the substitution then solve for y you should get a solution to the original equation.
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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?
Looking at the graph it is easy to see that y(0) = y_0 is equal to 1 which is also our C value
We can approximate b by using the point (-1, 0.5)
y = e^(-b t)
0.5 = e^(-b (-1))
0.5 = e^(b) take ln of both sides
ln(0.5) = ln(e^(b))
b = - 0.69
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Good.
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You are a little off on most of your solutions, but you have nearly all the pieces to make everything work.
Hopefully my notes will help to get you completely on track.
Once more, questions are welcome.
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