Query_19

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course Mth 279

3/31 13

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

Question: Find the general solution of the equation

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

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Your solution:

The complimentary solution is:

r^2 + 1 = 0

r = +/- i

The general solution is :

y(t) = c1 e^(i t) + c2 e^(-i t)

The real part of this is the complimentary solution:

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

The particular solution is:

y_p(t) = C e^t sin(t) + D e^t cos(t)

y_p’’(t) = 2 C e^t sin(t) - 2 D e^t sin(t)

Pluggin y_p(t) and y_p’’(t) into y’’ + y = e^t sin(t) we have

2C e^t cos(t) - 2D e^t sin(t) + Ce^t sin(t) + De^t cos(t) = e^t sin(t)

2 C cos(t) - 2 D sin(t) + C sin(t) + D cos(t) = sin(t)

(2 C + D) cos(t) + (C - 2D) sin(t) = sin (t)

(2 C + D) must = 0

(C - 2D) must = 1 for the above equation to be correct.

2 C + D = 0

C - 2D = 1

4C + 2D = 0

C - 2D = 1

Adding these together we have

5C = 1

C = 1 / 5

2 (1 / 5) + D = 0

D = - 2 / 5

Our particular solution is

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

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

y(t) = A cos(t) + B 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:

Characteristic equation:

r^2 + r = 0

r(r + 1) = 0

r = 0 , - 1

y_c(t) = A e^(r1 t) + B e^(r2 t)

y_c(t) = A + B e^-t

y_p(t) = A t^2 + Bt + C

We need to go one higher power to yield a t^2 term

y_p(t) = A t^3 + Bt^2 + Ct + D

y_p’(t) = 3 A t^2 + 2 B t + C

y_p’’(t) = 6 A t + 2 B

Plugging y_p’(t) and y_p’’(t) into y’’ + y’ = 6 t^2 we have:

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

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

3A = 6

6A + 2B = 0

2B + C = 0

A = 2

B = - 6

C = 12

y_p(t) = 2 t^3 -6 t^2 + 12 t + D = 0

The constant D is negligible since no y term (only y’ and y’’) appear in the equation.

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

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

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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:

The characteristic equation is

r^2 + r = 0

r(r +1) = 0

r= 0 , -1

y_c(t) = A e^(r1 t) + B e^ -t

y_c(t) = A + B e^-t

We expect the particular equation to be in the form of:

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

y_p’(t) = - C sin(t) + D cos(t)

y_p’’(t) = - C cos(t) - D sin(t)

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

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

D - C = 1

-D - C = 0

D = C + 1

-C - 1 - C = 0

2C = -1

C = - 1 / 2

D = 1 / 2

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

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

y(t) = A + b e^-t - 1 / 2 cos(t) + 1 / 2 sin(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:

USING METHOD OF UNDETERMINED COEFFICIENTS

y’’ - 2 y’ + 3 y = 0

We can set up a characteristic equation using the above homogeneous equation.

r^2 - 2r + 3 = 0

This gives two complex roots r1 and r2 in the form of r = alpha + beta i

r1 = 1 + i sqrt(2)

r2 = 1 - i sqrt(2)

This gives our general solution in the form of:

y(x) = c1 e^(alpha x) cos (beta x) + c2 e^(alpha x) sin(beta x)

alpha = 1

beta = sqrt (2)

First I will break down the 2 e^-t cos(t) term.

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

Second, I will break down the t^2 term

Yp(x) = C x^2 + D x + E

With E = 0 in this case

Last, I will break down the t e^(3t) term.

Yp(x) = F t e^(3t)

As this is duplicated in the initial equation, it needs to me multiplied by an additional factor of the independent variable (x). This leads to

Yp(x) = F t^2 e^(3t)

Adding these three above cases together we have:

Yp(x) = A e^-t sin(sqrt(2) ) + B e^-t cos(sqrt(2) x) + C x^2 + D x + F 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’’ + 4 y = 0

r^2 + 4 = 0

r = +/- 2i

alpha = 0 , beta = 2

y_c(x) = c1 e^(alpha x) cos(beta x) + c2 e^(alpha x) sin(beta x)

y_c(x) = c1 cos(2x) + c2 sin(2x)

The first term 2 sin(t)

Yp(x) = A sin(t) *

*The derivative will involve cos as well so

Yp(x) = A sin(t) + B cos(t)

The second term cosh(t)

@&

The cosh terms are for hyperbolic cosines. You should have seen them in first-year calculus. If not, you should study them. It's very easy to catch on to their derivatives.

cosh^2(t) = ((e^t + e^-t) / 2)^2.

The derivative of cosh(t) is sinh(t) (which is equal to (e^t - e^-t) / 2), and the derivative of sinh(t) is cosh(t).

*@

Yp(x) = C sin(t) + D cos(t)

The third term cosh^2 (t)

I am confused about how to break down this final term

Yp(x) = A sin(t) + B cos(t) + C sin(t) + D cos(t) + ???

This can be simplified to

Yp(x) = (A + C) sin(t) + (B + D) cos(t) + ????

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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:

The complementary equation has roots r = alpha + beta i

y_c(x) = c1 e^(alpha x) cos(beta x) + c2 e^(alpha x) sin(beta x)

With the given complimentary solution, alpha = 0 and beta = 1

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

As B sin(t) forms part of the complementary solution, this must be multiplied by an additional factor of t so

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

y_p’(t) = A + (B sin(t) + C cos(t)) + t(B cos(t) - C sin(t) )

y_p’’(t) = B cos(t) - C sin(t) + (B cos(t) - C sin(t) + t( -B sin(t) - C cos(t))

y_p’’(t) = 2B cos(t) - 2C sin(t) - t B sin(t) - t C cos(t)

The coefficient of the B and C terms in negligible because they are constants so

y_p’’(t) = B cos(t) - C sin(t) - t B sin(t) - t C cos(t)

Now, plugging y_p(t) , y_p’(t) and y_p’’(t) into the equation we have

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

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

A = 1

C = - 1

B = 0

y_p(t) = t - C t cos(t) )

y = c1 + c2 cos(t) + t - C t cos(t)

@&

Good.

I believe you can evaluate C. You'll get 1/2.

*@

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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:

y_p = A e ^ ( i 2 t)

y_p’ = 2 i A e ^ ( i 2 t)

y_p’’ = 4 i^2 A e ^ ( i 2 t) = - 4 A e ^ ( i 2 t)

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

-4 A - A = 1

A = - 1 / 5

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

y_p = - 1 / 5 (cos(2 t) + i sin(2 t))

Expanding the original equation we have:

y’’ - y = cos(2 t) + i sin(2 t) with the real part being

y’’ - y = cos(2 t)

The real part of the particular solution

y_p = - 1 / 5 cos(2 t)

y_p’ = 2 / 5 sin(2 t)

y_p’’ = 4 / 5 cos (2 t)

Substituting this into the real part of expanded original equation we have

( - 1 / 5 cos (2 t) )’’ - (- 1 / 5 cos(2 t) = cos(2t)

4 / 5 cos (2 t) + 1 / 5 cos(2 t) = cos(2 t)

This is correct.

Substituting the imaginary part of the expanded original equation we have

y_p = - 1 / 5 sin(2 t)

y_p’ = - 2 / 5 cos(2 t)

y_p’’ = 4 / 5 sin( 2 t)

4 / 5 sin(2 t) + 1 / 5 sin (2 t) = sin(2 t)

This is correct.

??? (I’ve ignored the i terms because they cancel out. Should I include them?

@&

Good, but be sure you are aware that both the real and imaginary parts of a solution are solutions to the equation.

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Given Solution:

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