Query 19

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course MTH 279

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

Question: Find the general solution of the equation

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

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Your solution:

We’ll start with the complementary solution.

y” + y = 0

λ^2 + 1 = 0

λ = ±i

y_1 = e^(it)

y_2 = e^(-it)

y_g = Ae^(it) + Be^(-it)

Using Euler’s formula:

y_g = Acos(t) + Bsin(t) + Acos(t) - Bsin(t)

y_c = 2Acos(t)

Now for the particular solution, we will use the method of undetermined coefficients.

Since the inhomogeneous term is e^t*sin(t), we can guess that the particular solution will take a form of y_p = Ae^(αt)sin(βt) + Be^(αt)cos(βt).

We take the first and second derivatives of y_p, and plug them into our original DE.

We get (2A+B)e^(t)cos(t) + (A-2B)e^(t)sin(t) = e^(t)sin(t).

Solving for A and B yields A = 1/5 and B = -2/5

So our particular solution is y_p = 1/5e^(t)sin(t) - 2/5e^(t)cos(t).

Our general solution is a combination of y_c and y_p.

y_g= 2Acos(t) + 1/5e^(t)sin(t) - 2/5e^(t)cos(t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Find the general solution of the equation

y '' + y ' = 6 t^2

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Your solution:

Complementary solution:

characteristic polynomial:

λ^2 + λ = 0

λ = -½ ± ½ = -1,0

y_1 = e^(-t)

y_2 = e^(0)

y_c = Ae^(-t) + B

Particular solution:

guess that solution will take the form y_p = Ct^3 + Dt^2 + Et

y_p’ = 3Ct^2 + 2Dt + E

y_p” = 6Ct + 2D

Plug back into the DE:

3Ct^2 + (6C+2D)t + 2D + E = 6t^2

3Ct^2 = 6t^2

C = 2

6C+2D = 0

D = -6

2D + E = 0

E = 12

y_p = 2t^3 - 6t^2 + 12t

y_g = Ae^(-t) + B + 2t^3 - 6t^2 + 12t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Find the general solution of the equation

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

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Your solution:

Complementary solution:

λ^2 + λ = 0

λ = -½ ± ½ = -1,0

y_1 = Ae^(-t)

y_2 = Be^(0)

y_c = Ae^(-t) + B

Particular solution:

Guess that the solution will come in form y_p = Csin(t) + Dcos(t).

y_p’ = Ccos(t) - Dsin(t)

y_p” = -Csin(t) - Dcos(t)

-Csin(t) - Dcos(t) + Ccos(t) - Dsin(t) = cos(t)

(C-D)cos(t) - (C+D)sin(t) = cos(t)

C-D = 1

C = 1+D

-C-D = -1-D-D = cos(t)

D = -½

C = ½

y_p = 1/2*sin(t) - ½*cos(t)

y_g = Ae^(-t) + B + ½*sin(t) - ½*cos(t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

The expected form would be:

y_p = Ae^(-t)sin(t) + Be^(-t)cos(t) + Ct^3 + Dt^2 + Et + Ft*e^(3t)

@&
Note that e^(3 t) is a solution of the homogeneous equation, which modifies the trial solution.

Also you don't need a t^3 term but you do need multiples of t and constants.
*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I think this would be correct for this situation. I think you would add the different cases, e^(αt)cos(βt), polynomial, and t*e^(αt).

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Self-critique rating: OK

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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_p = Asin(t) + Bcos(t) + Ccosh(t)^3 + Dcosh(t)^2 + Ecosh(t)

@&
Hyperbolic functions are just combinations of corresponding exponential functions, so you could get away with multiples of e^t, e^(-t), e^(2 t) and e^(-2 t), along with a plain constant term that results from the square of cosh(t).
*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I’m not entirely sure about this one. I’m not great with hyperbolic functions, and I couldn’t find any tables with this form listed. I know that the inhomogeneity is a sin or cos function, then we have Asin(βt) + Bcos(βt), but I’m not positive about the hyperbolic functions. I treated them as if they were a polynomial though, since one is squared and one is not. I’m not sure if this is possible.

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Self-critique rating: 3

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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:

λ = -α/2 ± sqrt(α^2 - 4β)/2

In order to have y_c = Acos(t) + Bsin(t), we must have complex roots, which means that 4β>α^2.

λ = -α/2 ± i*sqrt(4β - α^2)/2

y_1 = e^(α/2*t)(cos(β_2t) + i*sin(β_2t))

y_2 = e^(α/2*t)*(cos(β_2t) - i*sin(β_2t))

We know that cos(β_2t) = cos(t) and sin(β_2t) = sin(t), so β_2 = 1.

sqrt(4β - α^2)/2 = 1

Since y_c is not multiplied by an exponential function, we can determine that α = 0.

Which means that β = 1.

y” + y = t + sin(t)

y_p = At^2 + Bt + C*t*sin(t) + D*t*cos(t)

y_p’ = 2At + B + Ctcos(t) + Csin(t) - Dtsin(t) + Dcos(t)

y_p” = 2A + Ccos(t) - Ctsin(t) + Ccos(t) - Dtcos(t) - Dsin(t) - Dsin(t)

Plugging in, simplifying, and solving, we get:

A = 0, B = 1, C = 0, D = -1/2

y_p = t - tcos(t)/2

y_g = y_c + y_p = Acos(t) + Bsin(t) + t - tcos(t)/2

confidence rating #$&*:

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

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

y_p = Ae^(2it)

y_p’ = -2Asin(2t) + 2A*i*cos(2t)

y_p” = -4Acos(2t) - 4A*i*sin(2t)

-5Acos(2t) - 5A*i*sin(2t) = cos(2t) + i*sin(2t)

A = -1/5

y_ p = -e^(2it)/5 = -cos(2t)/5 - i*sin(2t)/5

y_p1 = -Acos(2t)/5

y_p1’ = 2Asin(2t)/5

y_p1” = 4Acos(2t)/5

4Acos(2t)/5 + Acos(2t)/5 = cos(2t)

5Acos(2t)/5 = cos(2t)

Acos(2t) = cos(2t)

A = 1

y_p2 = -i*sin(2t)/5

y_p2’ = -2i*cos(2t)/5

y_p2” = 4*i*sin(2t)/5

4*i*sin(2t)/5 + i*sin(2t)/5 = -i*sin(2t)

5*i*sin(2t)/5 = -i*sin(2t)

Both y_p1 and y_p2 are solutions of the differential equation.

confidence rating #$&*:

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

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK"

Query 19

@& Note that e^(3 t) is a solution of the homogeneous equation, which modifies the trial solution. Also you don't need a t^3 term but you do need multiples of t and constants. *@

@& Hyperbolic functions are just combinations of corresponding exponential functions, so you could get away with multiples of e^t, e^(-t), e^(2 t) and e^(-2 t), along with a plain constant term that results from the square of cosh(t). *@

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

@&
Note that e^(3 t) is a solution of the homogeneous equation, which modifies the trial solution.

Also you don't need a t^3 term but you do need multiples of t and constants.
*@

@&
Hyperbolic functions are just combinations of corresponding exponential functions, so you could get away with multiples of e^t, e^(-t), e^(2 t) and e^(-2 t), along with a plain constant term that results from the square of cosh(t).
*@