Query_14 

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course Mth 279

3/20 15

Query 14 Differential Equations*********************************************

Question: Decide whether y_1 = 3 e^t, y_2 = e^(t + 3) are solutions to the equation y '' - y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (-1) = 1 and y ' (-1) = 0.

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Your solution:

y_1(t) = 3 e^t

y_1’(t) = 3 e^t

y_1’’(t) = 3 e^t

y’’ - y = 0

3 e^t - 3 e^t = 0

0 = 0

y_2 = e^(t + 3)

y_2’= e^(t + 3)

y_2’’= e^(t + 3)

y’’ = y = 0

e^(t + 3) - e^(t + 3) = 0

0 = 0

Computing the wronksian

W(t) = [f(t) g(t)] = f(t) g’(t) - g(t)f’(t)

W(t) = 3 e^t e^(t + 3) - e^(t + 3) 3 e^ t = 0

Since W(t) = 0 , they are not a fundamental set and are linearly independent.

???? I am confused as to how to find the general solution.

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The Wronskian being zero implies that the functions are not linearly independent.

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y’’ - y = 0 We can write this using lambda (L) as

L^2 - 1 = 0 This factors to

(L - 1) (L + 1) = 0

Lambda = -1 , 1

this gives the general solutions as

y (t) = c_1 e^-t + c_2 e^ t

y’(t) = -c_1 e^-t + c_2 e^ t

y’(-1) = c_2 / e - c_1 e = 0

c_1 e + c_2 / e = 1

2 c_2 / e = 1

c_2 = e / 2

c_1 = 1 / 2 e

This gives the particular solution of

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

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Good, though you didn't have to find this.

My fault about the terminology, but this isn't a particular solution. We could call this a 'specific solution', but it's not a 'particular solution'. A particular solution is a solution to the nonhomogeneous equation. I used the word inappropriately in my statement of the problem.

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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y_1(t) = e^-t

y_1’(t) = -e^-t

y_1’’(t) = e^ - t

y’’ + 2 y’ + y = 0

e^-t - 2 e^-t + e^-t = 0

0 = 0

y_2(t) = 2e^(1-t)

y_2’(t) = -2e^(1-t)

y_2’’(t) = 2e^(1-t)

y’’+2 y’ + y = 0

2e^(1-t) - 4 e^(1-t) + 2 e^(1 - t) = 0

0 = 0

Computing the wronksian

W(t) = [f(t) g(t)] = f(t) g’(t) - g(t) f’(t)

W(t) = (e^-t) (-2 e^(1-t)) - (2 e ^(1 - t) ) (-e^-t) = 0

W(t) = 0

Since wronksian = 0 , they are not a fundamental set and therefore linearly dependent.

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Question: Suppose y_1 and y_2 are solutions to the equation

y '' + alpha y ' + beta y = 0

and that y_1 = e^(2 t). Suppose also that the Wronskian is e^(-t).

What are the values of alpha and beta?

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Your solution:

y_1(t) = e ^ (2 t)

y_1’(t) = 2 e ^ (2 t)

y_1’’(t) = 4 e ^ (2 t )

y’’ + A y’ + B y = -

4 e ^ (2 t) + A 2 e ^ (2t) + B e ^ (2 t)

e^(2t) ( 4 + 2 A + B ) = 0

Equation 1: 4 + 2 A + B = 0

The wronksian = e^ -t

W(t) = f(t) g’(t) + g(t) f’(t)

e^ 2t y_2’(t) + y_2(t) 2 e^ ( 2 t) = e^ -t

int [y_2(t)e^(2t)]’ = int [e^ -t]’

y_2(t) e^(2t) = -e^-t

y_2(t) = -e^(-3t)

y_2’(t) = 3 e^(-3t)

y_2’’(t) = -9 e^(-3t)

y’’ + Ay’ + By = 0

- 9 e^(-3t) + A 3 e^(-3t) + B e^(-3t) = 0

e^(-3t) (- 9 + 3 A - B) = -

Equation 2: 3 A - B - 9 = 0

We can use Equation 1 and Equation 2 to solve for A

Adding Equation 1 and Equation 2 together we have

2 A + B + 4 = 0

+

3A - B - 9 = 0

5 A - 5 = 0

A = 1

Plugging A = 1 into equation 1 we can solve for B

2A + B + 4 = 0

2(1) + B + 4 = 0

b = -6

So A = 1 and B = - 6

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Good.

Here's an alternative way to approach the first part of the problem:

The Wronskian is

det ( [ e^(2 t), y_2; 2 e^(2 t), y_2 ' ] = e^(-t)

Thus

y_2 ' e^( 2 t ) - 2 y_2 e^(2 t) = e^(-t).

Multiplying both sides by e^(-2 t) we get

y_2 ' - 2 y_2 = e^(-3 t).

This is a nonhomogeneous linear equation which, if necessary, can be solved using integrating factor e^(- 2 t). However by inspection it is not difficult to see that a solution is

y_2 = - e^(-3 t) / 5.

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