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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: 

 divide by e^t

e(-y) y' = -t + sin(t)

Integrate

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

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The derivative of

-1/2 t^2 - cos(t)

is -t + sin(t), but the derivative of

-e^(-y) y

is (e^(-y) * y - e^(-y) ) y ', not just e^(-y) y '.

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Y = 1/2 e^(y) * t^2 + e^y cos(t) - Ce^(y)

Initial conditions

0 = 1/2(0) + e^(y) - C e(y)

0 = e^y (1-C)

C = 1

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

 

 

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: 

 3y^2 y' = 1- 2t

 integrate

y^3 y = t - t^2 + C

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Your left-hand side is not correct.

The equation becomes

3 y^2 dy = (1 - 2t) dt

An antiderivative of the left-hand side is y^3.

The derivative of y^3 with respect to t is 3 y^2 y '.

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y = (t-t^2 + C)/y^3

initial conditions

-1 = (0 + 0 + C)/y^3

C =- y^3

y = (t-t^2 - y^3)/y^3

 

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

 not 100% sure if I did that right. I tried following the example in the book

 

 

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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: 

 

 take the derivative(?)

3y^2 y' + 2t + cos(y) y' = 0

y' (3y^2 + cos(y) ) = 2t

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The equation

y' (3y^2 + cos(y) ) = 2t

means

(3 y^2 + cos(y) ) dy/dt = 2 t,

which is effectively rearranged to

(3 y^2 + cos(y) ) dy = 2 t dt

and then integrated.

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when t = 4_0

y_0 ( cos(y)) = 8

I'm not sure how to go from here. I think I need a refresher on implicit differentiation too.

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Do see the Calculus qa review exercises.

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confidence rating #$&*:

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Given Solution: 

 

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

 was I on kind of on the right track ???

 

 

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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: 

 y' / (y^2 + 2y + 1) = sin(t)

integrate

I don't know how to integrate 1/(y^2 + 2y + 1)

I thought I could do u sub, but then I'd get u = y^2 + 2y + 1 and du = 2y +2 and that's not helpful

then I thought of the power rule

(y^2 + 2y + 1)^(-1)

but then I'd end up with ln (y^2 + 2y + 1) / (2y+2)

how is this done???

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You made a couple of good tries.

Factor the denominator, which gives you a perfect square. Then do what should be an obvious u substitution.

In general, had the quadratic not factored into a perfect square, you would try to factor and use partial fractions, or alternatively complete the square. These are standard first-year calculus techniques, and you should review them.

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- y / (y+1) = -cos(t)

y = cos(t) * (y +1)

solution exists from (- infin., + infin.)

 

 

confidence rating #$&*:

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Given Solution: 

 

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

 

 

 just not sure about integrating still

 

 

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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: 

For me this is problem 30 in section 2.7

I'm not at all sure how to solve for this problem.

Just by looking at the graphs I think y' = y^3 is graph B, only because it looks like a x^3 equation. By the same reasoning I think y' = -y^2 is graph A and y' = y(4-y) is graph C

I should be able to solve these and then match with the graphs

y' = y^3

integrate

y = 1/4 y^4

 But that doesn't look like any of these graphs

I don't think I fully understand this question

???

 

 y' = -y^2

integrate

y = -1/3 y^3, this sorta looks like graph B, but it's facing the wrong way

y' = 4y - y^2

integrate

y = 2y^2 -1/3 y^3, this doesn't really look like any of the graphs either.

 

 

 

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According to the graphs in the book:

y'=-y^2 --> Graph C

Concave up, asymptote at t = -1 and approaching 0 as t-> inf

y'= y^3 --> Graph A

Concave up but flipped of A with y --> 0 as t --> -inf and an asymptote at t= 1/2

y'= y(4-y) --> Graph B

For one, by exclusion but also because slope of y'-->0 as y--> 4 and y' --> 0 as y --> -inf

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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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Self-critique (if necessary):

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Self-critique rating:

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Check my notes.

You are using appropriate techniques for solving the equations, but as I believe you recognize you are running into problems with integration.

The only remedy is practice on integration. There are numerous web resources.

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