query_05

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course MTH 279

Query 05 Differential Equations*********************************************

Question: 3.2.6. Solve y ' + e^y t = e^y sin(t) with initial condition y(0) = 0.

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Your solution:

We know the equation is separable. We solve for y' = -e^y (t-sin(t)). Next we simplify by distributing the negative sign.

Dividing both sides by e^y we simplify to e^-y (dy/dt) = -t + sin(t). Integrating with respect to t we get

(integral of) e^-y (y') dt = (integral of) (-t + sin(t)) dt. Next we evaluate the integrals to get: -e^-y = -(t^2)/2 - cos(t) + C.

Then we solve for y: y(t) = -log((t^2)/2 + cos(t) - C). Using the initial condition, we get y(0) = -log(1 - C) = 0. Therefore,

since the log(1) = 0, C = 0.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.

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Your solution:

First we solve for y'. y' = (-2t + 1) / (3y^2). Then by multiplying both sides by 3y^2 we get: 3y'(y^2)dt = -2t + 1.

Next we integrate: (integral of) 3 y' (y^2) dt = (integral of) -2t + 1 dt. Then evaluate the integrals to get: y(t)^3 = -t^2 + t + C.

Solving for y: y(t) = cube root (-t^2 + t + C). Imposing the initial condition y(0) = -1, we get: y(0) = cube root (-(0)^2 + (0) + C) = -1,

we find that C = -1.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.

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Your solution:

Taking the implicit derivative we get: y' = -(3y^2 + cos(y)) / 2t. If an initial condition is at t = 2, the equation can be evaluated to be -(3y^2 + cos(y) / 4.

confidence rating #$&*:

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Given Solution: Not sure if I understood what I was looking for...not sure to work backward or forward.

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Self-critique (if necessary):

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Self-critique rating:

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Question: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) fand determine the t interval over which he solution exists.

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Your solution:

We know that this is a separable equation. The first thing we do s divide by (y^2 + 2y + 1) to get:

y' / (y^2 + 2y + 1) = sin(t). Next we integrate with respect to t to get (integral of) y' / (y^2 + 2y + 1) dt = (integral of) sin(t)dt.

Then we evaluate the integrals to get: -1/y + 1 = -cos(t) + C. Then we solve for y = (-cos(t) + C + 1) / (cos(t) - C).

Now we can find where t does not exist. Since we have cos(t) in the equation, we know when t = pi/2, 3pi/2, 5pi/2 etc it will cause cos(t)

to go to 0 and will make the bottom of the equation equal to 0 (and you cannot divide by 0). Therefore, t cannot equal +/-pi/2, +/-3pi/2, +/-5pi/2, etc

Confidence rating: 1

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Given Solution:

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Self-critique (if necessary):

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Question: 3.2.28. Match the graphs of the solution curves with the equations y ' = - y^2, y ' = y^3 an dy ' = y ( 4 - y).

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Your solution:

y ' = - y^2...the derivative: y = 1/C + x

y ' = y^3...the derivative: y = 3y^2

dy ' = y ( 4 - y)...the derivative: y = 4 - 2y

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I can't find the graphs for these solution curves? perhaps my browser isn't loading them properly

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Self-critique rating:"

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Good.

And finally you get to solve equations as separable.

You can also check out solutions at

http://vhmthphy.vhcc.edu/tests/shell_01/differential_equations/query_05_soln5749kjg0bnndjs9rfjfhjhe.htm

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