Query 05

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course MTH 279

11:17 am 2/28

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: 

e^-y dy= sin t -t dt 

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

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

y(0)= 0 --> C = 2

y = ln (2-cos t - 1/2t^2)

@&Almost, but

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

so

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

and

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

yielding C = 0

and solution

y = - ln | cos(t) + t^2 / 2 |*@

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 dy = 1-2t

y^3 = t-t^2 + C

y(0) = -1 ---> C = -1

 y = (t-t^2 - 1)^1/3

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: 

t = 2 --> y^3 + sin y = 0 --> y = 0 

y(2)= 0

Differentiating implicitly:

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

confidence rating #$&*:

@& Good.

Another route to reasoning out the solution:

This can be written as

y^3 + sin(y) = -t^2 + c,

which is the integral of the equation

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

so our equation is

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

To get the given implicit solution we need c = 4.

The left-hand side of the equation is equal to 0 when y = 0, and the right-hand side is 0 when t = sqrt(c) (provided c > 0).

Since we need c = 4, our condition

becomes y = 0 when t = sqrt(4) = 2,

which we express as

y(2) = 0.

*@

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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).

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Your solution: 

Should be y' = (y^2 + 2 y + 1 ) sin t?

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

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

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

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

@& should be (y+1) = cos(t) + c.

Next step should be

y = 1 / ( cos(t) + c ) - 1.*@

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

 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: 

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

confidence rating #$&*:

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&#Good responses. See my notes and let me know if you have questions. &#