QAs Feb 9

#$&*

course Mth 279

March 1 around 7:15pm.

q_a_05________________________________________

Differential Equations Notes

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

Question:

`q001. Solve the equation y ' = t^(3/2) tan(y) for the initial condition y(0) = pi/2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(dy/dt) = t^(3/2) tan(y)

Int (dy/tan(y)) = Int (t^(3/2) dt)

Int (cos(y)/sin(y) dy) = (2/5)t^(5/2) + c

Use “u” substitution:

u= sin(y) and du=cos(y) dy

ln |cos y| = (2/5)t^(5/2) + c

@& Should be ln | sin y |, not ln | cos y |. Otherwise good.*@

cos y = e^(2/5t^(5/2) + c)

cos y = Ae^(2/5t^(5/2)), A > 0

cos (pi/2) = Ae^(2/5t^(5/2))

1 = Ae^(2/5t^(5/2))

@&

tan(y) = sin(y) / cos(y).

Our equation is therefore defined only when cos(y) is not zero, i.e., when y is not equal to pi/2, 3 pi/2 or either of these angles plus a multiple of 2 pi.

Rearranging the equation we get

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

Integrating we get

ln (sin y)= 2/5 t^(5/2)

Solving for y:

y= arcSin(Ce^(2/5 t^(5/2)))

y(0) = 1 implies

pi / 2 = arcSin( A e^(2.5 * 0^(5/2) ) = arcSin(A)

so that

A = sin(pi/2) = 1.

The solution is therefore

y = arcSin(2/5 t^(5/2) ).

However there is a problem with this solution. y = pi/2 is not in the domain of definition of the function t^(3/2) tan(y).

*@

confidence rating #$&*:

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

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

Given Solution:

Self-critique:

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

Self-critique Rating:

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

Question: `q002. We saw previously that the equation y ' = -.01 t y^2 has solutions of the form y = 1 / (-.005 t^2 - c). For negative values of c this solution has a vertical asymptote at t = sqrt(200 c), and is not defined at t = sqrt(200 c).

We can understand why this equation has vertical asymptotes if we plot the direction field. Plot the direction field defined by t values 0, 2, 4, 6, 8 and 10, and y values 0, 10, 20, 30, 40. Explain what happens when you sketch solution curves from various points of this grid.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I plug in (t, y) into the y = 1 / (-.005 t^2 - c) function and solve for “c.” I can also plug in (t, y) into the y ' = -.01 t y^2 function to see the slope.

The slopes get “steeper” the further right and up on the graph you go. When “t” value is 0, all the slopes are 0.

confidence rating #$&*:

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

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

Given Solution:

`

Self-critique:

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

Self-critique Rating:

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

Question: `q003. The sort of thing that happened with the direction field of y = -.01 t y^2 cannot happen for a linear differential equation of the form y ' + p(t) y = 0, unless p(t) itself is undefined at a point. Explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y’ + p(t)y = 0 is a first order linear homogeneous differential equation.

It can be rewritten as y’ = -p(t)y (assuming that p(t) is continuous on the t-interval of interest)

After solving, y = e^(-P(t))

P(t) is an antiderivative of p(t) so y = e^(-Int p(t) dt)

p(t) is guaranteed to have an antiderivative since p(t) is continuous on (a,b).

P(t) = Int (p(s) ds, c, t)

The function P(t) is the particular antiderivative of p(t) that disappears at the point t=c.

@& The key is that e^(-x) is never negative, nor does it approach infinity at any positive value of x.*@

confidence rating #$&*:

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

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

Given Solution:

`

Self-critique:

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

Self-critique Rating:

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

Question:

`q004. Write the equation ( y^2 cos(t) ) ' = 0 in the form f(t, y, y') = 0. (To do so, find the derivative of y^2 cos(t) and set it equal to zero).

Show that this equation is solved by the function y for which y^2 cos(t) = c, where c is an arbitrary constant.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Derivative of: y^2 cos(t) wrt “t”: (-sin (t)*y^2) = 0

In the form f(t, y, y’): f(t, (y^2 cos(t)), (-sin(t)y^2)) Is this right?

How does (y^2 cos(t) = c) come in play when solving for y? I know there has to be another step before doing this.

@&

The equation becomes

2 y cos(t) y ' - y^2 sin(t) = 0

If y^2 cos(t) = c, then

y = sqrt( c / cos(t)).

y ' = -c sin(t) / cos^2(t) * 1 / (2 sqrt( c / cos(t) ) = c/2 sin(t) / cos^2(t) * sqrt(cos(t) / c) = sqrt(c) / 2 * sin(t) / cos^(3/2)(t)

where the function and its derivative are defined only if cos(t) > 0.

Substituting this function into the equation we get

2 sqrt(c / cos(t)) * sqrt(c) / 2 * sin(t) / cos^(3/2)(t) - c / cos(t) * sin(t) = 0.

Check to see that this simplifies to an identity.

`

*@

confidence rating #$&*:

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

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

Given Solution:

`

Self-critique:

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

Self-critique Rating:

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

Question: `q005. Show that the equation

dF = 0,

where F is a function F(t, y), is of the form

N(t,y) dt + M(t, y) dx = 0.

where N(t, y) = F_t and M(t, y) = F_y.

Show furthermore than N_y = M_t.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F is constant.

M = (dF/dt) = F_t

N = (dF/dy) = F_y

If Int (M dy) and Int (N dt) is “passed” then you can find the beginning function.

M_y = (F_t)_y = F_ty

N_t = (F_y)_t = F_yt

confidence rating #$&*:

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

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

Given Solution:

`

Self-critique:

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

Self-critique Rating:

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

Question:

`q006. Show that the equation

( sqrt(y) - e^t cos(y) + t ) dt + (t / (2 sqrt(y) + e^t sin(y)) dy = 0

is of the form

N dt + M dy = 0, with N_y = M_t.

****

Yes, IF:

N= ( sqrt(y) - e^t cos(y) + t )

M=(t / (2 sqrt(y) )+ e^t sin(y))

N_y=M_t?

Yes!

N_y=(1/(2 sqrt(y)) + e^t sin (y))

M_t=(1/(2 sqrt(y)) + e^t sin (y))

#$&*

&#This looks good. See my notes. Let me know if you have any questions. &#