Query 19

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course MTH 279

7/27/13 10:30pm

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

Question:  Find the general solution of the equation

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

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Your solution:

y"" + y = e^(t) sin(t)

let solution y = A e^(kt)

y' = k A e^(kt)

y"" = k^2 A e^(kt)

Plugging into equation we find:

k^2 + 1 = 0

k = i, -i

y_c = A e^(it) + Be^(-it)

y_c = C_1(cos(t) + i sin(t)) + C_2( cos(t) - i sin(t))

Assume:

y_p = A e^t sin(t) + B e^t cos(t)

y'_p = A e^t cos(t) + A e^t sin(t) - B e^t sin(t) + B e^t cos(t)

y""_p = 2 A e^t cos(t) - 2 B e^t sin(t)

plug y_p values into equation

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

factor

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

examining equation we see the following:

A - 2B = 1     and 2A + B = 0

We find:

A = 1/5,  B = -2/5

Thus

y = C_1(cos(t) + i sin(t)) + C_2(cos(t) - i sin(t)) + 1/5 e^t sin(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"" + y' = 6 t^2

y_c = c_1 + c_2 e^t

y_p = A t^3 + B t^2 + C t

y'_p = 3 A t^2 + 2 B t + C

y""_p = 6 A t + 2 B

plug y_p values into equation

6 A t + 2 B + 3 A t^2 + 2 B t + C = 6 t

3A = 6

6A + 2B = 0

2B + C = 0

A = 2

B = -6

C = 12

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

y(t) = c_1 + c_2 e^-t + 2t^3 - 6t^2 + 12t

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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"" + y' = cos(t)

y_c = c_1 + c_2 e^-t

y_p = A cos(t) + B sin(t)

y'_p = -A sin(t) + B cos(t)

y""_p = -A cos(t) - B sin(t)

-A cos(t) - B sin(t) + B cos(t) - A sin(t) = cos(t)

cos(t) (B-A) + sin(t) (-B - A) = cos(t)

B - A = 1

-B - A = 0

A = -1/2,  B = 1/2

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

y(t) = c_1 + c_2 e^-t + 1/2 sin(t) - 1/2 cos(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 '' - 2 y ' + 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t)

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Your solution:

 ****

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

 for 2e^(-t)

t^r [Ae^(alpha t)sin(t) + Be^(-t)cos(t)]

t^r[Ae^(1 - sqrt(2)i*t) + Bee^(1 - sqrt(2)i*t)

Eulers identity:

A[e^t cos(sqrt(2)t) + i e^t sin(sqrt(2)t)] + B[e^t cos(sqrt(2)t) + i e^t sin(sqrt(2)t)]

C_1(e^t cos(sqrt(2)t) + C_2 i e^t sin(sqrt(2)t)

C_1 e^t cos(sqrt(2)t) + C_2 e^t sin(sqrt(2)t)

y_p = A_1 e^(-t) sin(t) + A_2 e^(-t) cos(t) + B_1 t^2 + B_2 t + B_3 + (C_1 t + C_2) e^(3t)

@&

Since e^(3 t) is a solution to the homogeneous equation (solve the homogeneous equation carefully in order to see this) the term C_2 e^(3t) of (C_1 t + C_2) e^(3t) would give you zero, regardless the value of C_2. So this part of your trial solution gives you no information.

You would need to try the form

(C_1 t^2 + C_2 t) e^(3t)

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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"" + 4y = 2 sinh(t) cosh(t) + cosh^2 (t)

y_c = c_1 sin(2t) + c_2 cos(2t)

rewrite hyperbolic sign and cos using exponential

y"" + 4y = 2( (e^t - e^(-t)) /2) ((e^t + e^(-t))/2) + ((e^t + e(-t))/2)^2)

simplify

y"" + 4y = e^(2t) - e^(-2t) + 1

three parts to y_p

the form for the first part: A e^(2t)

second part: B e^(-2t)

third part : C

y_p = A e^(2t) + B e^(-2t) + C

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Given Solution:

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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:

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

y_c = c_1 cos(t) + c_2 sin(t)

alpha = 0,  beta = 1

y"" + y = t + sin(t)

I am not sure that I am right at this point. The particular solution for the t+ sin(t) part will have two parts with the first part (associate with the t) equal to a constant, and the second part (associated with sin(t)) being A sin(t) + B cos(t).  However, when I take the derivatives of the second part and plug them in, they add to zero.  I think I must have done something wrong.

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You have a good start.

The complementary solution results from a sum of multiples of e^(i t) and e^(-i t), which would result from the equation y '' + y = 0. So alpha = 0 and beta = 1.

Our equation is thus

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

It should be clear that if y = t then y '' + y = 0 + t = t, so the term t will be included in our particular solution.

sin(t) is a solution to the homogeneous equation, so our trial solution will include multiples of t sin(t) + t cos(t)

Our trial solution is therefore

y_P = t + A t sin(t) + B t cos(t).

Our derivatives are

y_P ' = 1 + A sin(t) + B cos(t) + t ( A cos(t) - B sin(t))

and

y_P '' = 2 A cos(t) - 2 A sin(t) + t ( -A sin(t) - B cos(t) ).

Substituting into our equation we get

y_P '' + y_P = t + sin(t)

2 A cos(t) - 2 B sin(t) + t ( -A sin(t) - B cos(t) ) + t + A t sin(t) + B t cos(t) = t + sin(t)

2 A cos(t) - 2 B sin(t) + t = t + sin(t)

The equation is satisfied if A = 0 and B = -1/2.

Thus

y_P = t - 1/2 * t cos(t)

and the general solution is

y(t) = c_1 cos(t) + c_2 sin(t) + t - 1/2 * t cos(t)

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Given Solution:

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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:

 ****

7.) y'' - y = e^(it2)

using A e^(i2t)

y = A e^(i2t)

y' = 2 i A e^(i2t)

y'' = -4 A e^(i2t)

-4 A e^(i2t) + A e^(i2t) = e^(i2t)

e^(i2t) * (-4A + A) - e^(i2t)

-4A + A = 1

A = -1/3

y_p = -1/3 e^(i2t)

y_p = -1/3(cos(2t) + i sin(2t))

y_p = -1/3 cos(2t) + 1/3 i sin(2t))

u(t) = -1/3 cos(2t) ----- v(t) = -1/3 i sin(2t)

u'(t) = 2/3 cos(2t) ----- v'(t) = -2/3 i sin(2t)

u''(t) = 4/3 cos(2t) ----- v''(t) = 4/3 i sin(2t)

4/3 cos (2t) - 1/3 cos(2t) = Real [e^(2 i t)] = cos(2t) + i sin(2t)

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Given Solution:  

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