Assignment 21

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course Mth 279

7/22

Query 19 Differential Equations*********************************************

Question: Find the general solution of the equation

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

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Your solution:

1) find y_c

r^2 +1 = 0

r = +- sqrt(-1)

r = +- i

y_1 = e^it y_2 = e^-it

Using Euler’s Formula

y_c = C_1sin(t) + C_2cos(t)

2) find y_p?

g(t) = e^tsin(t)

choosing a form

y_p = Ae^tsin(t) + Be^t*cos(t)

take derivatives

y’p = Ae^tcos(t) + Ae^tsin(t) + Be^t(-sin(t)) + Be^tcos(t)

y’’p = Ae^tsin(t) + Ae^tcos(t) + Ae^tcos(t) + Ae^tsin(t) - Be^tcos(t) - Be^tsin(t) + Be^tcos(t)

= 2Ae^tcos(t) - 2 Be^tsin(t)

Plugging into y’’ + y = e^tsin(t)

2Ae^tcos(t) - 2 Be^tsin(t) + Ae^tsin(t) + Be^t*cos(t) = e^t sin(t)

combine sin and cos to get system of eq:

A - 2B = 1

2A + B = 0 ; multiply by 2 to be rid of B

A - 2B = 1

4A + 2B = 0

5A = 1

A = 1/5

So B

1/5 - 2B = 1

-2B = 4/5

B = -2/5

y(t) = C_1sin(t) + C_2cos(t) + 1/5 e^tsin(t) + -2/5 e^t*cos(t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Will there be a table given on the test for y_p or is that table something to be memorized???

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You need to know what forms will work for various possible functions.

This is more a matter of practice and common sense than memorization.

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Not fully confident on how r is actually chosen? I know it can be 0,1, or 2 is it just trial and error??

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I'm not sure what r you are asking about.

The standard use of r in this context is for the coefficient of the exponent in the exponential function, which can take values found by solving the characteristic equation.

If you mean something else, can you clarify?

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Self-critique rating:

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Question: Find the general solution of the equation

y '' + y ' = 6 t^2

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Your solution:

y_c?

r^2 + r = 0

r(r + 1) = 0

r = 0 r = -1

y_c = C_1 e^(0t) + C_2e^-1t

find y_p?

g(t) = 6t^2

y_p = t[At^2 + Bt + C] ; r = 1

y_p = At^3 + Bt^2 + Ct

y’_p = 3At^2 + 2Bt + C

y’’_p = 6At + 2B

plugging in

6At + 2B + 3At^2 + 2Bt + C = 6t^2

3A = 6

6A + 2B = 0

2B + C = 0

A = 2

6(2) + 2B = 0

B = -6

2(-6) + C = 0

C = 12

Y(t) = C_1 e^(0t) + C_2e^-1t + 2t^3 + -6t^2 + 12t

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: Find the general solution of the equation

y '' + y ' = cos(t).

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Your solution:

y_c?

r^2 + r = 0

r(r+1) = 0

r = 0 r = -1

y_c = C_1 e^(0t) + C_2e^-1t

y_p?

g(t) = cos(t)

y_p = t^0[Asin(t) + Bcos(t)]

y_p = Asin(t) + Bcos(t)

y’_p = Acos(t) + -Bsin(t)

y’’_p = -Asin(t) - Bcos(t)

plugging in

-Asin(t) - Bcos(t) + Asin(t) + Bcos(t) = cos(t)

A - B = 1

-A - B = 0

so A = - B

-B-B = 1

-2B = 1

B = -1/2

so A = 1/2

y(t) = C_1 e^(0t) + C_2e^-1t+ 1/2 sin(t) + -1/2 cos(t)

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: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:

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

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Your solution:

y_c?

r^2 - 2r - 3 = 0

(r-3)(r+1) = 0

r = 3 r =-1

y_c = C_1e^(3t) + C_2e^(-1t)

y_p?

g(t) = 2e^(-t)cos(t) + t^2 + t^e^(3t)

take each part separately

2e^(-t)cos(t) -> y_p = Ae^-tcos(t) + Be^-tsin(t)

t^2 -> y_p = Ct^2 + Dt + E

te^(3t) -> (Ft^3 + Gt^2 + Ht + I)e^(3t)

y(t) = C_1e^(3t) + C_2e^(-1t) + Ae^-tcos(t) + Be^-tsin(t) + Ct^2 + Dt + E + (Ft^3 + Gt^2 + Ht + I)e^(3t)

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Pretty good, but you probably have a couple of unnecessary terms in your trial solution.

The complementary solution is A e^(3 t) + B e^(-t)

The term 2 e^-t cos(t) is expected to result from a combination of multiples of e^-t cos(t) and e^(-t) sin (t).

The term t^2 is expected to result from a second-degree polynomial of the form a t^2 + b t + c, with a = 1. The constants b and c can be adjusted so that the various derivatives of the other terms cancel.

The term t e^(3 t) would be expected to result from a combination of multiples of t e^(3 t) and e^(3 t), except that e^(3 t) is a solution of the homogeneous equation. So we will try a combination of t^2 e^(3 t) and t e^(3 t).

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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: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:

y '' + 4 y = 2 sin(t) + cosh(t) + cosh^2(t).

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Your solution:

y_c?

r^2 + 4 = 0

r^2 = -4

r = +- 2i

y_c = C_1sin(2t) + C_2cos(2t)

y_p?

g(t) = 2sin(t) + cosh(t) + cosh^2(t)

2sin(t) -> Asin(t) + Bcos(t)

cosh(t) = e^t/2 + e^-t/2 -> Ce^t + De^-t

cosh^2(t) -> (e^t/2 + e^-t/2)^2 = e^2t/2 + e^-2t/2 -> Ee^(2t) + Fe^(-2t)

y_p = Asin(t) + Bcos(t) + Ce^t + De^-t + Ee^(2t) + Fe^(-2t)

y(t) = C_1sin(2t) + C_2cos(2t) + Asin(t) + Bcos(t) + Ce^t + De^-t + Ee^(2t) + Fe^(-2t)

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cosh^2(t) = (e^(2 t) + 2 + e^(-2 t)) / 2 and will result from a sum of multiples of e^(2 t), e^(-2 t) and 1.

So your trial solution needs to include a constant, to cover the need for a multiple of 1.

Otherewise very good.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: The equation

y '' + alpha y ' + beta y = t + sin(t)

has complementary solution y_C = c_1 cos(t) + c_2 sin(t) (i.e., this is the solution to the homogeneous equation).

Find alpha and beta, and solve the equation.

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Your solution:

1) find a and b?

r^2 + ar + b = 0

a = 0 b/c y_c is in the format of having r be imaginary that way we are left with

r^2 + b = 0

r^2 = -b

r = +- sqrt(b)i

There b= 1 b/c there isn’t a number in front of the t

2) Solve

t -> y_p = At + B

sin(t) -> y_p = t(Csin(t) + Dcos(t))

y_p = At + B + Ctsin(t) + Dtcos(t)

y’_p = A + Ctcos(t) + Csin(t) + Dt(-sin(t)) + Dcos(t)

y’’ = -Ctsin(t) + Ccos(t) + Ccos(t) - Dtcos(t) - Dsin(t) + D(-sin(t))

plugging in

-Ctsin(t) + 2Ccos(t) - Dtcos(t) - 2Dsin(t) + At + B + Ctsin(t) + Dtcos(t) = t + sin(t)

2Ccos(t) - 2Dsin(t) + At + B = t + sin(t)

2C = 0 -> C = 0

-2D = 1 -> D= -1/2

1A = 1 -> A = 1

B = 0

y(t) = C_1cos(t) + C_2sin(t) + t + -1/2t cos(t)

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: Consider the equation

y '' - y = e^(`i * 2 t),

where `i = sqrt(-1).

Using trial solution

y_P = A e^(i * 2 t)

find the value of A, which is in general a complex number (though in some cases the real or imaginary part of A might be zero)

Show that the real and imaginary parts of the resulting function y_P are, respectively, solutions to the real and imaginary parts of the original equation.

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Your solution:

A) y_p = Ae^(2ti) ; y’’ - y = e^(i(2t))

y’_p = 2iAe^(2it)

y’’p = -4Ae^(2it)

plugging in

-4Ae^(2it) + Ae^(2ti) = e^(2it)

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This would be

-4Ae^(2it) - Ae^(2ti) = e^(2it),

which is what you probably used since the next equation is correct.

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divide by e^(2it)

-4A - 1A = 1

-5A = 1

A = -1/5

y_p = -1/5e^(2it)

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Right, but you want to note that

y_P = -1/5 e^(i * 2t) = -1/5 ( cos(2t) + i sin(2 t))

for use below.

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B) Not really sure what the second question is asking us to find???

so I found what y_c is??

y’’ - y = 0

r^2 -1 = 0

r = +- 1

y_c = C_1 e^(t) + C_2e^(-t)

y(t) = C_1 e^(t) + C_2e^(-t) - 1/5e^(2it)

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Expanding the exponential to show its real and imaginary parts you get the equation

y '' - y = cos(2 t) + i sin(2 t)

The real part of this equation is

y '' - y = cos(2 t)

and the real part of the solution found above is -1/5 cos(2t). If you plug this solutoin into the real part of the equation it works. This is what you were asked to show.

You were also asked to show the same thing about the imaginary part of the original equation and the imaginary part of its solution.

This illustrates the very important principle that the real and imaginary parts of a general solution to an equation satisfy the respective real and imaginary parts of that equation.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

07-23-2014

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