Query_19

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Question: Find the general solution of the equation

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

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Your solution:

y(t) = y_c(t) + y_p(t)

Starting with the solution of the homgeneous: y_c(t)

P(a) = a^2 + 1 = 0

Since this does not split into its roots, we must use the Quradtatic Equation.

(-b +- sqrt(b^2 - 4ac)) / 2a

a = 1

b = 0

c = 1

This yeilds a solution: a_1,2 = +- i

due to the neagative under the radiacal: sqrt(0^2 - 4(1)(1)) = sqrt(-4)

y1(t) = e^(i*t)

y2(t) = e^(-i*t)

Using Euler's Formula yeilds:

y_c(t) = Asin(t) + Bcos(t)

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Next we need the other part: y_p(t)

since g(t) = e^t sin(t)

we can choose the particular solution form:

y_p(t) = t^r (Ae^(at) sin(bt) + Be^(at) cos(bt))

r = 0 since neither would use the same form as the y_c(t) solution.

y_p(t) = Ae^(t) sin(t) + Be^(t) cos(t)

If we plug this into the original differential equation, we can solve for the unknown constants A, and B.

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

(Ae^(t) sin(t) + Be^(t) cos(t))'' + (Ae^(t) sin(t) + Be^(t) cos(t)) = e^t sin(t)

(Ae^(t) cos(t) + Ae^(t) sin(t) - Be^(t) cos(t) + Be^(t) cos(t))' + ...

-Ae^(t) sin(t) + Ae^(t) cos(t) + Ae^(t) cos(t) + Ae^(t) sin(t) - Be^(t) cos(t) - Be^(t) sin(t) - Be^(t) sin(t) + Be^(t) cos(t) + Ae^(t) sin(t) + Be^(t) cos(t) = e^t sin(t)

Simplifying due to some terms cancel when added yeilds:

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

Now we can get a series of equations to solve the constants:

for e^t sin(t): A - 2B = 1

for e^t cos(t): 2A + B = 0

multipling the second by 2 yeilds: 4A + 2B = 0

A - 2B = 1

4A + 2B = 0

adding the 2 above equations, causes b to cancel yeilding:

5A = 1

A = (1/5)

Pluging this back into one of the equations, we can find B.

A - 2B = 1

(1/5) - 2B = 1

-2B = (4/5)

B = (-2/5)

Therefore y_p(t) = (1/5)e^t sin(t) - (2/5)e^t cos(t)

and since y(t) = y_c(t) + y_p(t)

y(t) = Asin(t) + Bcos(t) + (1/5)e^t sin(t) - (2/5)e^t cos(t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: Find the general solution of the equation

y '' + y ' = 6 t^2

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Your solution:

y(t) = y_c(t) + y_p(t)

Starting with the solution of the homgeneous: y_c(t)

P(a) = a^2 + a = 0

which splits into the roots: a(a + 1)

a1 = 0

a2 = -1

And from these roots we get to solutions:

y1(t) = e^(-t)

y2(t) = e^(0) = 1

Resulting in a solution:

y_c(t) = Ae^(-t) + B

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Next we need the other part: y_p(t)

since g(t) = 6t^2

we can choose the particular solution form:

y_p(t) = t^r (At^2 + Bt + C )

r = 1 since one would use the same form as the y_c(t) solution.

y_p(t) = At^3 + Bt^2 + Ct

If we plug this into the original differential equation, we can solve for the unknown constants.

y '' + y ' = 6 t^2

(At^3 + Bt^2 + Ct)'' + (At^3 + Bt^2 + Ct)' = 6 t^2

(3At^2 + 2Bt + C)' + 3At^2 + 2Bt + C = 6t^2

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

Rearranging this yeilds:

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

Now we can get a series of equations to solve the constants:

for t^2: 3A = 6

for t: 6A + 2B = 0

for c: 2B + C = 0

Solving the first equation:

3A = 6

A = 2

Plugging in this for the other 2 equations:

6A + 2B = 0

6(2) + 2B = 0

2B = -12

B = -6

2B + C = 0

2(-6) + C = 0

-12 + C = 0

C = 12

Therefore y_p(t) = 2t^3 - 6t^2 + 12t

and since y(t) = y_c(t) + y_p(t)

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

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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(t) = y_c(t) + y_p(t)

Starting with the solution of the homgeneous: y_c(t)

P(a) = a^2 + a = 0

which splits into the roots: a(a + 1)

a1 = 0

a2 = -1

And from these roots we get to solutions:

y1(t) = e^(-t)

y2(t) = e^(0) = 1

Resulting in a solution:

y_c(t) = Ae^(-t) + B

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Next we need the other part: y_p(t)

since g(t) = cos(t)

we can choose the particular solution form:

y_p(t) = Asin(t) + Bcos(t)

If we plug this into the original differential equation, we can solve for the unknown constants A, and B.

y '' + y ' = cos(t)

(Asin(t) + Bcos(t))'' + (Asin(t) + Bcos(t))' = cos(t)

(Acos(t) - Bsin(t))' + Acos(t) - Bsin(t) = cos(t)

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

Now we can get a series of equations to solve the constants:

for sin(t): -> -A-B = 0

for cos(t): -> A-B = 1

Adding these 2 together yeilds:

-2B = 1

B = (-1/2)

Plugging in this for the other equation:

A-(-1/2) = 1

A + (1/2) = 1

A = (1/2)

Therefore y_p(t) = (1/2)sin(t) - (1/2)cos(t)

and since y(t) = y_c(t) + y_p(t)

y(t) = Ae^(-t) + B + (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(t) = y_c(t) + y_p(t)

Starting with the solution of the homgeneous: y_c(t)

P(a) = a^2 + -2 + 3 = 0

Since this does not split into its roots, we must use the Quradtatic Equation.

(-b +- sqrt(b^2 - 4ac)) / 2a

a = 1

b = -2

c = 3

This yeilds a solution: a_1,2 = 1 +- i*sqrt(2)

due to the neagative under the radiacal, we get imaginary solutions.

y1(t) = e^(1 + i*sqrt(2)t)

y2(t) = e^(1 - i*sqrt(2)t)

Using Euler's Formula yeilds:

y_c(t) = e^t (Asin(sqrt(2)t) + Bcos(sqrt(2)t))

Next we need the other part: y_p(t)

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

this must be done in parts

u(t) = 2 e^-t cos(t)

v(t) = t^2

w(t) = te^(3 t)

y_p(t) = u(t) + v(t) + w(t)

choosing the forms to use:

u(t) = Ae^(-t)sin(t) + Be^(-t)cos(t)

v(t) = Ct^2 + Dt + E

w(t) = (Ft + G)e^(3 t)

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

@&
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 need to try a combination of t^2 e^(3 t) and t e^(3 t).
*@

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

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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(t) = y_c(t) + y_p(t)

Starting with the solution of the homgeneous: y_c(t)

P(a) = a^2 + 4 = 0

Since this does not split into its roots, we must use the Quradtatic Equation.

(-b +- sqrt(b^2 - 4ac)) / 2a

a = 1

b = 0

c = 4

This yeilds a solution: a_1,2 = +- 2*i

due to the neagative under the radiacal, we get imaginary solutions.

y1(t) = e^( i*2t)

y2(t) = e^(- i*2t)

Using Euler's Formula yeilds:

y_c(t) = Asin(2t) + Bcos(2t)

Next we need the other part: y_p(t)

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

this must be done in parts

but first cosh(t) must be converted into e^(t) form.

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

g(t) = 2 sin(t) + (e^t/2 + 1/(2e^t)) + (e^t/2 + 1/(2e^t))^2

g(t) = 2 sin(t) + (1/2)e^t + (1/2)e^(-t) + (1/4)e^(2t) + (1/4)e^(-2t) + (1/2)

choosing the forms to use:

y_p1(t) = Asin(t) + Bcos(t)

y_p2(t) = Ce^t

y_p3(t) = De^(-t)

y_p4(t) = Ee^(2t)

y_p5(t) = Fe^(-2t)

y_p6(t) = G

since y_p(t) = y_p1(t) + y_p2(t) + y_p3(t) + y_p4(t) + y_p5(t) + y_p6(t)

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

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: 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:

Due to the Quadratic Equation

(-b +- sqrt(b^2 - 4ac)) / 2a

And Euler's Formula:

y_c(t) = e^(at)(Asin(wt) + Bcos(wt))

Since our solution is basically:

y_c(t) = Asin(t) + Bcos(t)

Alpha must be zero, otherwise there would be an e^(at) in our homogeneous solution.

b in our quadratic equation is alpha in Euler's formula, so b = 0

and sqrt(-4ac) must then equal sqrt(-4), so that sqrt(4) = 2, and the bottom can be cleared, which is 2(a) = 2

making a = 1, and c = 1

yeilding a solution of +,- i

Therefore our equation should be

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

with Alpha = 0, and Beta = 1

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y_p(t) = u(t) + v(t)

u(t) = At + B

v(t) = t^r(Csin(t) + Dcos(t))

r = 1 because otherwise it would be the same as the homogeneous solution.

v(t) = Ctsin(t) + Dtcos(t)

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

If we plug this into the original differential equation, we can solve for the unknown constants

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

(At + B + Ctsin(t) + Dtcos(t))'' + (At + B + Ctsin(t) + Dtcos(t)) = t + sin(t)

(A + Ctcos(t) + Csin(t) - Dtsin(t) + Dcos(t))' + At + B + Ctsin(t) + Dtcos(t) = t + sin(t)

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

Simplifying.

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

Now we can get a series of equations to solve the constants:

for cos(t): -> 2C = 0

for sin(t): -> -2D = 1

for t: -> A = 1

and for c: -> B = 0

Solving all the equations:

2C = 0

C = 0

-2D = 1

D = -1/2

A = 1

B = 0

y_p(t) = t + (-1/2)tcos(t)

y(t) = Asin(t) + Bcos(t) + t - (1/2)tcos(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:

Using: y_P = A e^(i * 2 t)

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

(A e^(i * 2 t))'' - (A e^(i * 2 t)) = e^(i * 2 t)

(2A*i e^(i * 2 t))' - A e^(i * 2 t) = e^(i * 2 t)

-4Ae^(i * 2 t) - A e^(i * 2 t) = e^(i * 2 t)

For the equation to work:

-4A -A = 1

-A5 = 1

A = -1/5

y_p(t) = (-1/5)e^(i * 2 t)

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now solving for the homogneous solution of the equaiton

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

y'' - y = 0

P(a) = a^2 - 1 = 0

(a - 1)(a + 1) = 0

a1 = 1

a2 = -1

P_c(t) = Ae^(t) + Be(-t)

y(t) = Ae^(t) + Be(-t) - (1/5)e^(i * 2 t)

with the homogeneous solution being the real part, and the particular solution is the imaginary part.

confidence rating #$&*:

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Given Solution:

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Query_19

@& 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 need to try a combination of t^2 e^(3 t) and t e^(3 t). *@

&#Your work looks very good. Let me know if you have any questions. &#

@&
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 need to try a combination of t^2 e^(3 t) and t e^(3 t).
*@