Query_05

#$&*

course Mth 279

3/18Been forgetting to send this one for a while now, I have had it finished.

Query 05 Differential Equations

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

dy/dt = e^y(sin(t) - t)

int(dy/e^y) = int(sin(t) - t)dt

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

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

y(0) = 0

y(0) = -ln((1/2)(0)^2 + cos(0) - c)

e^0 = -cos(0) -c

1 = -1-c

2 = -c

c = -2

@&

Since ln(1) = 0,

-ln((1/2)(0)^2 + cos(0) - c) = 0

means that

((1/2)(0)^2 + cos(0) - c) = 1

so that

1 - c = 1

which gives us c = 0.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This time I tried the equation form N(y) + M(t) = C

3 y^2 y ' + 2 t = 1

int(3 y^2 y ') + int(2 t -1)dt = 0

@&

You would write this

int(3 y^2 dy) + int(2t-1 dt).

int(3 y^2 y ' ) would mean int (3 y^2 dy/dt), which isn't quite right.

*@

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

y^3 = c - t(t - 1)

y = Cube root (c - t(t-1))

y(o) = -1

y(0) = Cube root (c - 0(0-1)) = -1

cube root (c) = -1

C = -1^3 = -1

@&

Good.

Compare with the following, which is completely equivalent:

This equation is easily rearranged into the form

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

which we integrate to get

y^3 = t - t^2 + c.

Solving for y we get

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

Applying the initial condition y(0) = -1 we have

-1 = (0 - 0^2 + c)^(1/3)

so that c = -1.

Our final solution is

y = (t - t^2 - 1) ^ (1/3).

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I believe that this problem has to be worked backwards.

y^3 + t^2 + sin(y) = 4

y^3 + sin(y) = 4 - t^2

d/dy ( y^3 + sin(y)) = d/dt(4 - t^2)

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

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

Checking the intial condition.

y(2) = 0

0 = 4 - t^2

0 = 4 - 2^2

0 = 4 - 4

0 = 0

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y ' = (y^2 + 2 y + 1) sin(t)

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

int(y+1)^-2 dy = -cos(t) + c

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

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

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

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

The Solution may exist everywhere but where:

cos(t) + c = 0

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

2

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Originally I had solved each of these but I can figure out which graph they represent with their slopes.

y ' = - y^2

As y increases, y ' is a negative slope that increases by the square of y

making it graph C

y ' = y^3

As y increases, y' is the cube root of y, so its slope is positive

making it graph A

y ' = y ( 4 - y)

As y incrases, y' is the more complex equation above, but its good to note that by pluging in zero or 4 like on the graph for y we get a slope of zero.

This matches our last graph which is B

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

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