Query 05

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course Mth 279

3/18 8

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:

I'm doing this by separation now, because I think this is where you said i'd have to learn how to do it, and it is going well.

first solve for dy(t)/dt = e^y(t)*(sin(t) - t)

divide by e^y(t)

e^-y(t)*dy(t)/dt = sint - t

now integrate each side wrt t

-e^-y(t) = -cos(t) - t^2/2 + c

take the ln of each side to solve for y(t), being careful with the signs

y(t) = -log(cos(t) + t^2/2 - c)

now take y(0) = 0, solve for c, to find that c = 0

now if we put c = 0 into our y(t) equation, we find that its just the same equation, just take the c off, because its 0.

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

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Your solution:

solve for dy(t)/dt, bring 3y(t)^2 back over to make easier to integrate wrt t

left with y(t)^3 = t-t^2 + c

y(t) = cubed root ( t - t^2 + c) { going to substitute the cubed root as CR }

now sub our y(0) = -1 in

-1 = CR ( -1 -1^2 + c)

-1 = CR c

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:

i'm not completely positive on what you mean on the state a problem, so i worked backwards until I got a problem, which I guess could be stated in a way.

working bakcwards to find an answer for c

sub in 2 for t to find y^3 + sin(y) = 0

integrate wrt t

y(t) ^4/4 - cos(y) = c

solve for c with the initial condition of t = 2 to find that c = 4 - cosy

sub back into the old equation

y(t)^4 / 4 - cos(y) = 4 - cos(y)

solve this equation with the intitial condition of y(0) = 2

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Not bad but you're going in the wrong direction.

If you had a differential equation in y and t you would effectively integrate.

Here you have the solution, which is effrectively the integral of the differential equation.

If you differentiate the implicit solution, in other words, you get the equation.

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

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Your solution:

??????????

Ok, I'm not sure if this can be solved without knowing the intial conditions. y(0) = ?

anyways, just solving for c, get to an integration

dy(t)/dt sin(t)^-1 dt = ( y(t)^2 + 2y(t) + 1) dt

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You have y's on the right-hand side with dt and t's on the left with dy.

Get the y's with the dy and the t's with the dt and then integrate.

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y(t)*csc(t) = y(t)^3 + y(t)^2 y(t) + c

where c = y(t)*( csc(t) - y(t)^2 - y(t) - 1 )

here i'm not sure where to go, or if i'm supposed to know an intial condition for y(t) to solve for c, or to have some kind of number value,

Every example problem in the book has initial conditions to go by.

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: 3.2.28. 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:

I'm not completely sure what to do with this, so I solved for y by integrating to get a constant for each equation.

now solving for c to get these three equations

c_1 = y + y^3/3

c_2 = y - y^4/4

c_3 = y^3/3 - 2y^2 + y

Now using values of -2, -1, 0, 1, and 2 I substituted these values in for y in our c equations, to find c

I graphed each c equation and the overall slope in the first quadrant is about 2/3

When I say overall slope, i mean the ""solution curve"" for each combined graph are all basically a slope of 2/3

each graph seems to come from the left, intersecting at the origin, and going in the positive x and y direction until the point 1,1 where the c_3 graph then falls into negative y numbers

confidence rating #$&*:

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You can solve these equations but that's not the point here. You need to learn what to do if you can't solve them.

You need to analyze the behavior of each function and use that behavior to sketch your slope field.

For example - y^2 is always negaitive unless y = 0, in which case it's zero. As the magnitude of y increases the magnitude of y^2 increases at an increasing rate. These behaviors are then translated into the behavior of the slope field.

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Given Solution:

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Self-critique (if necessary):

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Check my notes, and check the site

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

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