Query 19

#$&*

course Mth 279

7/28 1:34 pm

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

Question:  Find the general solution of the equation

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

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 solving for yc

y^2 + 1 = 0

y1 = i

yc = c1 cos(t) + c2 sin(t)

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

yp' = A e^(t) sin(t) + A e^(t) cos(t) + Be^(t) cos(t) - Be^(t) sin(t)

yp'' = A e^(t) sin(t) + A e^(t) cos(t) + A e^(t) cos(t)- A e^(t) sin(t) + Be^(t) cos(t) -Be^(t) sin(t) - Be^(t) sin(t) - Be^(t) cos(t) = 2Ae^(t) cos(t) - 2B e^(t) sin(t)

plugging into initial equation

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

A - 2B = 1

2A + B = 0

A = -1/5

B = -2/5

y = yc + yp

y = c1 cos(t) + c2 sin(t) -2/5 e^(t) cos(t) - 1/5 e^(t) sin(t)

 

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

Question:  Find the general solution of the equation

y '' + y ' = 6 t^2

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 y^2 + y = 0

y = 0,-1

yc = c1 + c2 e^(-t)

 using yp = At^2 does not work because the t^2 term would not appear in the second and first derivative. I must multiply by t

yp = At^3 + Bt^2 + Ct

yp' = 3At^2 + 2Bt + C

yp'' = 6At + 2B

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

3A = 6

A = 2

B = C = 0

yp = 2t^3

y = c1 + c2 e^(-t) + 2t^3

 

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

Question:  Find the general solution of the equation

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

y^2 + y = 0

y(y+1) = 0

y1 = 0, y2 = -1

yc = c1 + e^(-t)

yp = Asin(t) + Bcos(t)

yp' = Acos(t) - Bsin(t)

yp'' = -Asin(t) - Bcos(t)

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

-A - B = 0

A - B = 1

B = -1/2

A = 1/2

yp = 1/2 sin(t) - 1/2 cos(t)

y = c1 + e^(-t) + 1/2 sin(t) - 1/2 cos(t)

 

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 using superposition

2 e^-t cos(t)

yp_1 = Ae^(-t)sin(t) + Be^(-t) cos(t)

t^2

yp_2 = Ct^2 + Dt + E

t e^(3 t)

y_3 = Fte^(3t)

yp = Ae^(-t)sin(t) + Be^(-t) cos(t) + Ct^2 + Dt + E + Fte^(3t)

@&

You need to solve the homogeneous equation first, to be sure none of the terms on the right-hand side coincide with solutions of that equation.

It turns out that e^(3 t) is a solution of the homogeneous equation, so you would need to use

(F t^2 + G t) e^(3t)

in your trial solution.

*@

 

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 Should I memorize what cosh simplifies to??

using superposition

2 sin(t)

 yp_1 = Asin(t) + Bcos(t)

cosh(t) = 1/2 e^(t) + 1/2 e^(-t)

yp_2 = Ce^(t)

yp_3 = De^(-t)

cosh^2(t) = [1/2 e^(t) + 1/2 e^(-t)]^2 = 1 + e^(2t)/4 + e^(-2t)/4

yp_4 = F

yp_5 = Ge^(2t)

yp_6 = He^(-2t)

yp = Asin(t) + Bcos(t) + Ce^(t) + De^(-t) + F + Ge^(2t) + He^(-2t)

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 because cos and sin are in the complimentary solution, that means that y1 and y2 are complex numbers. Because there is no e term, alpha is 0, and beta is 1 because the t in cos(t) has a coefficient of 1

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

finding yp using superposition

t

yp_1 = At

yp_1' = A

yp_1'' = 0

0 + At = t

A = 1

sin(t)

using yp_2 = Asin(t) + Bcos(t) does not work

yp_2 = Atsin(t) + Btcos(t) 

 yp_2' = Asin(t) + Atcos(t) + Bcos(t) -Btsin(t)

yp_2'' = Acos(t) + Acos(t) - Atsin(t) -Bsin(t) -Bsin(t) - Btcos(t) = 2Acos(t) - Atsin(t) - 2Bsin(t) -Btcos(t)

2Acos(t) - Atsin(t) - 2Bsin(t) -Btcos(t) + Atsin(t) + Btcos(t) = sin(t)

2Acos(t) -2Bsin(t) = sin(t)

But I can't solve for two variables with one equation ???

 

@&

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)

*@

 

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 y^2 - 1 = 0

y1 = e^(t) y2 = e^(-t)

yc = c1 e^(t) + c2 e^(-t)

yp = Ae^(i 2t)

 yp' = i 2A e^(i 2t)

yp'' = -4A e^(i 2t)

 -4A e^(i 2t) - Ae^(i 2t) = e^(i 2t)

-4A - A = 1

A = -1/5

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

 

 

 Although I'm not sure what you mean by the real and imaginary part of the solution. If yp = Asin(2t) + Bcos(2t), you would have separate parts of the solution (one imaginary and one real). ???

 

 

@&

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.

If y_P = A e^(i * 2 t) then

y_P ' = 2 i A e^(i * 2 t) and

y_P '' = -4 A e^(i * 2 t).

Substituting this into the original equation we get

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

so that

-4 A - A = 1

and

A = -1/5.

This isn't complex, as was advertised in the statement of the problem. However in general we do expect A to be complex; in this case the imaginary part of the solution was zero.

In any case the particular solution is

y_P = -1/5 e^(i * 2t) = -1/5 ( cos(2t) + i sin(2 t)).

Expanding the right-hand side of the original equation, using Euler's identity, we have

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

y '' - y = cos(2 t) + i sin(2 t)

The real part of the original equation is

y '' - y = cos(2 t)

and the real part of the particular solution is -1/5 cos(2 t). Substituting this for y we get

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

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

which is clearly true.

The imaginary part of the original equation is

y '' - y = sin(2 t)

and the imaginary part of the particular solution is -1/5 sin(2 t). Substituting this for y we get

(-1/5 sin(2 t) ) '' - (-1/5 sin(2 t) ) = sin (2 t)

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

which is clearly true.

As noted previously, the solution for A did turn out to have imaginary part zero. This is not generally the case; it happened here because there was no y ' component in the equation.

Had the original equation been, for example,

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

then substitution of y = A e^(i * 2 t) would yield

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

so that

(-4 + 2 i - 2) A = 1

and A = 1 / (-6 + 2 i) = (-6 - 2 i) / 40 = -3/20 - i / 20.

In this case A e^(i * 2 t) would be -(3/20 - i / 20) ( cos(2 t) + i sin(2 t) ) = (-3 cos(2 t) + sin( 2 t) ) / 20 + i ( -3 sin(2 t) - cos(2 t) ) / 20.

A e^(i * 2 t) = -(3/20 - i / 20) ( cos(2 t) + i sin(2 t) )

= (-3 cos(2 t) + sin( 2 t) ) / 20 + i ( -3 sin(2 t) - cos(2 t) ) / 20.

You can verify that the real part

(-3 cos(2 t) + sin( 2 t) ) / 20

is a solution to the real part of the original equation, which is

y '' + y ' - 2 y = cos(2 t).

You can also verify that the imaginary part

( -3 sin(2 t) - cos(2 t) ) / 20

is a solution to the imaginary part

y '' + y ' - 2 y = sin(2 t)

of the original equation.

*@

confidence rating #$&*:

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

.............................................

Given Solution: 

 

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

Self-critique (if necessary):

 

 

------------------------------------------------

Self-critique rating:

*********************************************

Question: 

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

________________________________________

#$&*

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