Assignmet 21

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course Mth 279

7/28

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)

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Given Solution:

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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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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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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)

@&

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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Given Solution:

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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)

@&

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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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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)

divide by e^(2it)

-4A - 1A = 1

-5A = 1

A = -1/5

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

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)

@&

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