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course MTH 279
Solve each equation:1. y ' + y = 3
e^-t
-e^-t * y ' + e^-t *y = 3 e^-t
S(y* e^-t)’ dt = S (3 e^-t) dt
y*e^-t = -3t * e^-t + C
y = -3t + C/ e^-t
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C / e^-t = C e^t.
y ' would be -3 + C e^t, so our equation would become
-3 + C e^t + 3t + C e^t = 3.
So your solution doesn't work.
You should have multiplied both sides by integrating factor e^t, not by e^-t.
This would have led to the correct solution, which is
y = 3 + c e^(-t).
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Part of the solution process is always to check your solution by plugging it into the original equation. This will quickly give you the feedback you need to correct errors.
In this case the integral of the coefficient of y, which is the integral of 1 with respect to t, is t so your integrating factor would be e^t.
This is to be contrasted with the solution of y ' + y = 0, which is y = C e^-t, for reasons that should be clear from the preceding assignment.
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2. y ' + t y = 3 t
e^(-t^2/2)
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This would be a solution to the equation y ' + t y = 0, but it isn't the correct integrating factor for the given nonhomogeneous equation.
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-t e^(-t^2/2) * y’ + t e^(-t^2/2) * y = 3t * e^(-t^2/2)
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The left-hand side is not the differential of y e^(-t^2 / 2).
Had you used integrating factor e^(t^2 / 2), the differential of y e^(t^2 / 2) would have been e^(t^2 / 2) * y ' + t y e^(t^2 / 2), which is identical to the product of the original left-hand side and the integrating factor.
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S(y* e^(-t^2/2))’ dt = S(3t * e^(-t^2/2)) dt
U = -(t^2)/2
Du = -t dt
y* e^(-t^2/2) = -1/3 (e^(-t^2/2)) + C
y = -1/3 + C/ e^(-t^2/2)
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This solution will not check out. See also my preceding note.
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3. y ' - 4 y = sin(2 t)
e^ (4t)
4e^(4t) y’ - (4e^(4t)y) = sin(2t)*(e^(4t))
S (e^(4t) * y)’ dt = S ( sin(2t) * (e^(4t))) dt
e^(4t) * y = S e^(4t) sin(2t) dt
Integrate by parts:
u = e^(4t) dv = sin(2t) dt
du = 4 e^(4t) dt v = -1/2 cos(2t)
S e^(4t) sin(2t) dt = -1/2 e^(4t) cos(2t) + S 2e^(4t) cos(2t) dt
Integrate by parts again:
u = e^(4t) dv = 2 cos(2t) dt
du = 4 e^(4t) dt v = sin(2t)
S e^(4t) sin(2t) dt = -1/2 e^(4t) cos(2t) + e^(4t) sin(2t) - S 4 e^(4t) sin(2t) dt
S e^(4t) sin(2t) dt + S 4 e^(4t) sin(2t) dt = -1/2 e^(4t) cos(2t) + e^(4t) sin(2t)
5 S e^(4t) sin(2t) dt = 1/2 e^(4t) (-cos(2t) + 2 sin(2t))
intgrl e^(4t) sin(2t) dt = 1/10 e^(4t) (-cos(2t) + 2 sin(2t)) + C
e^(4t) * y = 1/10 e^(4t) (-cos(2t) + 2 sin(2t)) + C
y = 1/10 ((-cos(2t) + 2 sin(2t)) + C / e^(4t)
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4. y ' + y = e^t, y (0) = 2
e^(-t)
- e^(-t) y’ + e^(-t) y = e^(t) * e(-t)
S (e^(-t) *y)’ dt = S dt
e^(-t) *y = t + C
y = (t + C) / e^(-t)
y = (t+ C)e^(t)
y(0) = 2
2 = (0 + C)e^(0)
C = 2
(t+ 2)e^(t)
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5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2
e^ (-3t)
-3 e^ (-3t) y’ + 3 e^ (-3t) y = 3 e^ (-3t) + e^t * e^ (-3t) + 2t e^ (-3t)
S (e^ (-3t) * y) ‘ dt = S (= 3 e^ (-3t) + e^t * e^ (-3t) + 2t e^ (-3t)) dt
e^ (-3t) * y = -e^-3t - ½ e^(-2t) - 2/3t e^ (-3t) - 2/3 e^ (-3t) + C
y = -1 - ½ e^t - 2/3 t - 2/3 + C
y = - ½ e^t - 2/3 t - 5/3 + C
y(1) = e^2
e^2 = ½ e - 2/3 - 5/3 + C
e^2 - ½ e + 7/3
y = e^2 - ½ e + 7/3 + ½ e^t - 2/3 t - 5/3
y = e^2 - 2/3t + 2/3 - ½ e + ½ e^t
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6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)?
p(t) = 2t because the integral of 2t is t^2
g(t) = e^(t^2) because e^0 is 1
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i don't believe any of your solutions check out, nor are the left-hand sides of the equations you get when you multiply by your proposed integrating factors equal to the necessary differentials. Check my notes on the first couple of problems.
An easy adjustment will lead to correct solutions. However you should always check that the left-hand side is the differential of y * integrating factor, and you should always check your solution by substituting it into the original equation.
If you do those two things you will understand the method, avoid errors, and be fully prepared when equations of this form appear on tests.
You should rework and resubmit this exercise.
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i don't believe any of your solutions check out, nor are the left-hand sides of the equations you get when you multiply by your proposed integrating factors equal to the necessary differentials. Check my notes on the first couple of problems.
An easy adjustment will lead to correct solutions. However you should always check that the left-hand side is the differential of y * integrating factor, and you should always check your solution by substituting it into the original equation.
If you do those two things you will understand the method, avoid errors, and be fully prepared when equations of this form appear on tests.
You should rework and resubmit this exercise. I don't believe you'll find it difficult or overly time-consuming to do so, and it will be well worth your effort.
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