Query 02

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Your solution:

y’ = -y + 3

y’ / (y + 3) = 1 divide each side by y + 3

int y’ / (y + 3) = int 1 integrate each side

-ln (y + 3) = x + c_1

y = -e ^ (-x - c_1) + 3

y = c_1 e^-x + 3

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You should be solving these equations using the methods of the corresponding section of the text. This equation certainly is separable, but you want to treat it as a nonhomogeneous linear equation. You will need experience solving equations of this type.

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Given Solution:

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Question:

2. y ' + t y = 3 t

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Your solution:

y’ = 3 t - t y

y’ = t (-y + 3)

y’ / (-y + 3) = t

int y’ / (-y + 3) = int t

-ln (-y + 3) = t^2 / 2 + c_1

y = c_1 e^(t^2 / 2) + 3

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Given Solution:

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Question:

3. y ' - 4 y = sin(2 t)

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Your solution:

Integrating factor = e ^ int -4 dt = e^ -4 t

Multiply each side by integrating factor

(e^ -4 t) y’ - (4 e^-4 t) y = (e ^ -4 t) sin(2t)

Integrate both sides

int [(e^ -4 t) y’ - (4 e^-4 t) y] = int [(e ^ -4 t) sin(2t) ]

Reverse product rule states that the left side of the equation becomes the integrating factor multiplied by y.

(e^-4 t) y = (- 1 / 10 e ^ -4 t) (cos(2t)) + sin(2t)) + c_1

y = - 1 / 10 cos (2t) - 1 / 5 sin(2t) + c_1 e^4 t

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Good.

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Given Solution:

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Question:

4. y ' + y = e^t, y (0) = 2

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Your solution:

y’ = e^t - y

u(t) = e^t

(e^t)y’ + (e^t)y = e^2 t multiply each side by u(t)

e^t (dy(t) / dt) + (d / dt) e^t y(t) = e ^ 2 t

(d / dt) (e^t y(t)) = e^2t

int (d / dt) (e^t y(t)) = int e^2t integrate both sides

e^t y(t) = e^2t / 2 + c_1

y(t) = e^t / 2 + c_1 e^-t divide each side by e^t

this above is the general solution

Use the initial conditions to solve for c_1

2 = 1 / 2 + c_1

c_1 = 3 / 2

The solution is:

y(t) = 1 / 2 e^ -t (e ^ 2 t + 3)

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Given Solution:

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Question:

5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2

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Your solution:

p(t) = 3

P(t) = 3 t

Multiply each side by P(t)

e^(3t) y’ + 3 e^(3t) y = 3 e^(3t) + 2 e^(3t) t + e^(3t) e^t

(e^(3t) y)’ = 3 int e^(3t) + 2 int e^(3t) t + int e^(4t)

e^(3t) y = 3 e^(3t) + (e^(4t) / 4) + 2 / 9 e^(3t) (3 t - 1) + c e^(-3t)

y = 3 + 2 / 9 e^(3t) (3 t - 1) + 1 / 4 e^(4t)

This is the general solution. Plugging in y(1) = e^2 we have

e^2 = c e^-3 + 4 / 9 e^3 + e / 4 + 3

c = -3 e^3 - e^4 / 4 + e^5 - 4 e^6 / 9

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Good, but check details against the following:

y ' + 3 y = 3 + 2 t + e^t

y 'e^(3t) + 3 ye^(3t) = 3e^(3t) + 2 te^(3t) + e^(4t)

(ye^(3t))' = 3e^(3t) + 2 te^(3t) + e^(4t)

int ((ye^(3t))') = int (3e^(3t) + 2 te^(3t) + e^(4t))

ye^(3t) = e^(3t) + (2/9) e^(3t) (3t - 1) + (1/4)e^(4t) + c

y(t) = c e^(-3t) + (2/3)t + (1/4)e^t + (7/9)

y(1) = e^2

y(1) = c e^(-3) + (2/3) + (1/4)e + (7/9) = e^2

c = e^5 - (1/4)e^4 - (13/9)e^3

So

y(t) = (e^5 - (1/4)e^4 - (13/9)e^3) e^(-3t) + (2/3)t + (1/4)e^t + (7/9)

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Given Solution:

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Self-critique (if necessary):

I don’t believe I integrated properly. The algebra in finding c was difficult.

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Question:

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)?

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Your solution:

From the general solution, P(t) is -t^2.

p(t) is the derivative of P(t) so p(t) = -2t

Working backwards from y = c e(^-t^2) + 1 we multiple each side by e^(t^2).

e^(t^2)y’= e^(t^2) * (1) + C

taking the dervative of each side gives us

e^(t^2)y = 2te^(t^2)

g(t) is the beginning of the right side of the equation or 2t.

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Given Solution:

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Self-critique (if necessary):

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Self-critique (if necessary):

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#*&!

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I think you're in good shape here. Check my notes.

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