Query 17

#$&*

course Mth 279

April 30 around 4:40pm.

Query 17 Differential Equations*********************************************

Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I read the given solution and understand this. However, how do you get the repeated solution, which in this case was r = (-2/5)?

@&
Our characteristic equation is

25 r^2 + 20 r + 4 = 0

with repeated solution r = -2/5

yielding solution set

{e^(-2/5 t), t e^(-2/5 t) }

and general solution

y(t) = A e^(-2/5 t) + B t e^(-2/5 t)

satisfying

y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

These conditions lead to the equations

A e^-2 + 5 B e^-2 = 4 e^-2

-2/5 A e^(-2) - 2/5 * 5 B e^-2 + B e^-2 = -3/5 e^-2,

or dividing both equations through by e^-2, and multiplying through the second by 5

A + 5 B = 4

-2 A - 5 B = -3.

The solution is A = 1, B = 3/5, giving us the solution

y(t) = e^(-2/5 t) - 3/5 t e^(-2/5 t).

*@

confidence rating #$&*:

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

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

Given Solution:

Our characteristic equation is

25 r^2 + 20 r + 4 = 0

with repeated solution r = -2/5

yielding solution set

{e^(-2/5 t), t e^(-2/5 t) }

and general solution

y(t) = A e^(-2/5 t) + B t e^(-2/5 t)

satisfying

y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

These conditions lead to the equations

A e^-2 + 5 B e^-2 = 4 e^-2

-2/5 A e^(-2) - 2/5 * 5 B e^-2 + B e^-2 = -3/5 e^-2,

or dividing both equations through by e^-2, and multiplying through the second by 5

A + 5 B = 4

-2 A - 5 B = -3.

The solution is A = 1, B = 3/5

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The characteristic equation is

3 r^2 + 2 sqrt(3)r + 1 = 0

with repeated solution r = ? (on the previous question, I asked how to find the repeated solution, but ill solve what I can without it)

I can’t find the yielding solution set { } without knowing the repeated solution, right?!

The general solution:

y(t) = A e^(r) + B t e^(rt)

satisfying

y(0) = 2 sqrt(3) and y' (0) = 3

These conditions lead to the equations:

I would then plug in the “r” value into the y(t) equation above and solve for A and B.

@& The repeated solution leads to the fundamental set containing y_1 = e^(-sqrt(3) / 3 * t), and y_2 = t e^(-sqrt(3) / 3 * t).

The characteristic equation is

3 r^2 + 2 sqrt(3) r + 1 = 0

with repeated solution

r = (-2 sqrt(3) +- sqrt( 12 - 12 ) ) / 6 = -sqrt(3)/3.

This yields fundamental set

{e^(-sqrt(3) / 3 * t), t e^(-sqrt(3) / 3 * t) }

with general solution

y(t) = A e^(-sqrt(3) / 3 * t) + B t e^(-sqrt(3) / 3 * t).

y(0) = A = 2 sqrt(3)

y ' (0) = -sqrt(3) / 3 * A + B = 3.

Substituting A = 2 sqrt(3) into the second equation we get

-6 + B = 3

so that

B = 9.

Our solution is therefore

y(t) = 2 sqrt(3) e^(-sqrt(3) / 3 * t) + 9 t e^(-sqrt(3) / 3 * t).

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question: Solve the equation

y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0,

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Reduction of order: y2(t) = v(t)y1(t)

y2(t) = v*sin(t)

y2’(t) = v*cos(t)

y2’’(t) = v*(-sin(t))

Plug into main equation: y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0

= [v*(-sin(t))] - [2 cot(t)* (v*cos(t))] + (1 + 2 cot^2t)*(v*sin(t)) = 0

After rearranging and simplifying, I got:

= [(-2*(cos(t) - sin(t))*v] / [tan(t)]

I’m doing something wrong here, I think. It might be my derivatives at the beginning. Should I have done like v’ and v’’?!

Anyways, after you get this, you should do v(t) = int ( w dt) and choose the constants to be anything.

You will then get the second solution which will then let you find the general solution for this problem.

@& We begin by letting y_2(t) = u(t) * y_1(t), so that

y_2 ' = u ' * y_1 + u * y_1 ' = u ' sin(t) + u cos(t)

and

y_2 '' = u'' * y_1 + 2 u ' * y_1 ' + u y_1 ''

= u '' * sin(t) + 2 u ' * cos(t) - u * sin(t).

Our equation becomes

u'' * y_1 + 2 u ' * y_1 ' + u y_1 '' - 2 cot(t) (u ' * y_1 + u * y_1 ') + (1 + 2 cot^2(t)) * u y_1 = 0

which can be rearranged to yield

[ u y_1 '' - 2 cot(t) * u y_1 ' + (1 + 2 cot^2(t)) * u y_1 ] + u '' y_1 + u ' ( 2 y_1 - 2 cot(t) y_1) = 0.

The terms in brackets can be expressed as u ( y_1 '' - 2 cot(t) * y_1 ' +(1 + 2 cot(t)^2) y_1); since y_1 is a solution to our original equation these terms therefore add up to zero.

This leaves us with

u '' + 2 u ' y_1 (1 - cot(t)) = 0.

or substituting y_1 = sin(t)

u '' + 2 u ' sin(t) ( 1 - cot(t)) = 0.

cot(t) = cos(t) / sin(t), and our equation for u becomes

u '' + 2 sin(t) u' - 1 = 0.

Letting v = u ' our equation is

v ' + 2 sin(t) v = 1.

This is a first-order linear equation. We have thus reduced the order of our equation from 2 to 1.

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*