#$&*
course Mth 279
4/12 6I said something really stupid about the cross product being the same as the determinant on a previous assignment, please over look that comment. It was made at 3 or 4 in the morning.
Query 19 Differential Equations
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Question: Find the general solution of the equation
y '' + y = e^t sin(t).
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Your solution:
Complimentary solution is
y_c(t) = Acos(t) + Bsin(t)
Trail form for our Particular solution is
y_p(t) = e^t [Asin(t) + Bcos(t)]
Plugging into our equation
y '' + y = e^t sin(t)
(e^t [Asin(t) + Bcos(t)])’’ + e^t [Asin(t) + Bcos(t)] = e^t sin(t)
(2Ae^t cos(t) - 2Be^t sin(t)) + Ae^t sin(t) + Be^t cos(t) = e^t sin(t)
e^t cos(t)(2A+B) + e^t sin(t)(A - 2B) = e^t sin(t)
Now setting coefficients equal,
2A + B = 0
A - 2B = 1
A=1/5, B= -2/5, so that Particular solution is
y_p = e^t [(1/5)sin(t) - (2/5)cos(t)]
General solution is given by y(t) = y_c(t) + y_p(t), so we have
y(t) = [Acos(t) + Bsin(t)] + e^t [(1/5)sin(t) - (2/5)cos(t)]
confidence rating #$&*:
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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: Find the general solution of the equation
y '' + y ' = 6 t^2
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Your solution:
Charcteristic equation gives us r = -1 and 0
Complimentary solution is
y_c(t) = c1e^-t + c2e^(0*t) = c1e^-t + c2
Trail form for our Particular solution is
y_p(t) = At^2 + Bt + C
Plugging into our equation
y '' + y’ = 6t^2
(At^2 + Bt + C)’’ + [At^2 + Bt + C]’ = 6t^2
(2A) + (2At + B) = 6t^2
Now setting coefficients equal,
2A + B = 0
A - 2B = 1
???I’ve messed up somehow going to try again?????
Trail form for our Particular solution is
y_p(t) = t[At^2 + Bt + C] = At^3 + Bt^2 + Ct
Plugging into our equation
y '' + y’ = 6t^2
(At^3 + Bt^2 + Ct)’’ + [At^3 + Bt^2 + Ct]’ = 6t^2
( 6At + 2B) + (3At^2 + 2Bt + C) = 6t^2
Now setting coefficients equal,
3A = 6
6A + 2B = 0
2B + C = 0
A= 2, B= -6, C= 12 so that Particular solution is
y_p = 2t^3 - 6t^2 + 12t
General solution is given by y(t) = y_c(t) + y_p(t), so we have
y(t) = c1e^-t + c2 + 2t^3 - 6t^2 + 12t
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
So the reason I ended up with my coefficient’s equation is due to the fact that I did not have my t^r in my equation, but my question is in the book they said to make t^r with r as small as possible so that no term in the assumed form is a solution of the homogeneous equation ay’’ + by’ + cy = 0, where r will be 0,1, or 2. What exactly does this mean?????
@& If you tried a t^2 + b t + c the last term c would be redundant with the constant term c_2 in your solution.
So you try t ( a t^3 + b t + c).
That works.
If it didn't then you would try t^2 ( a t^2 + b t + c), then t^3 ( ...), etc.. until you got to a power that did work.*@
Also the quadratic formula gave me r = -1 and 0, when r = 0 would I leave c2 in the equation or if the e^r_2*t term is not present do we drop the c2.
c1e^(-1*t) + c2e^(0*t) = c1e^(-t) + c2, or is it simply
c1e^(-1*t) + c2e^(0*t) = c1e^(-t)
???????????????????????????????????
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Self-critique rating:
@& e^0 = 1 so the solution is as you gave it.
The left-hand side is y '' - y ', so y = c_2 is certainly a valid solution, and must be included.
You did a good job on this solution. In fact from your solution I discovered an error in mine.*@
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Question: Find the general solution of the equation
y '' + y ' = cos(t).
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Your solution:
Again we have r = -1 and 0
Complimentary solution is
y_c(t) = c1e^(-t) + c2????I’m pretty sure this is correct????
Trail form for our Particular solution is
y_p(t) = Acos(t) + Bsin(t)
Plugging into our equation
y '' + y ' = cos(t).
(Acos(t) + Bsin(t))’’ + (Acos(t) + Bsin(t))’ = cos(t)
(-Acos(t) - Bsin(t)) + (-Asin(t) + Bcos(t)) = cos(t)
cos(t)(B - A) + sin(t)(-A - B) = cos(t)
Now setting coefficients equal,
-A + B = 1
-A - B = 0
A= -1/2, B= 1/2, so that Particular solution is
y_p(t) = (-1/2)cos(t) + (1/2)sin(t)
General solution is given by y(t) = y_c(t) + y_p(t), so we have
y(t) = [c1e^(-t) + c2] - (1/2)cos(t) + (1/2)sin(t)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
Same concern about the constant c2 in the complimentary solution.
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@& For reasons very similar to those on the preceding, y = c2 is a solution to the equation and must be included.
The fundamental set is
{e^-t, 1}.*@
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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:
Knowing…
If y_p1(t) is a particular solution for
y’’+ p(t)y’ + q(t)y = g_1(t)
and if y_p2(t) is a particular solution for
y’’+ p(t)y’ + q(t)y = g_2(t)
Then y_p1(t) + y_p2(t) is a particular solution for
y’’+ p(t)y’ + q(t)y = g_1(t) + g_2(t)
We would simply find the form of 2 e^-t cos(t), t^2 and t e^(3 t)
So for y '' - 2 y ' = 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t)
y_p(t) = [e^t (Asin(t) + Bcos(t))] + [t(At^2 + Bt + C)] + [e^3t(At + B)]
Confidence rating:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& There's a typo in the equation as I gave it. Should be
y '' - 2 y ' + 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t).
Having 3 y on the left side produces a non-constant solution:
The complementary solution is A e^(3 t) + B e^(-t)
The term 2 e^-t cos(t) is expected to result from a combination of multiples of e^-t cos(t) and e^(-t) sin (t).
The term t^2 is expected to result from a second-degree polynomial of the form a t^2 + b t + c, with a = 1. The constants b and c can be adjusted so that the various derivatives of the other terms cancel.
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 will try a combination of t^2 e^(3 t) and t e^(3 t).
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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(t) = [Acos(t) + Bsin(t)] + [e^t(At + B)] + [e^-t(At + B)] + [e^2t(At + B)] + [e^-2t(At + B)]
???Not sure and without working out just thinking that because cosh(t) = [(e^t + e^-t)/2] and cosh^2(t) = [(e^2t + e^-2t)/2]?????????????????
confidence rating #$&*:
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@& I haven't worked it out to verify, but the complementary solution doesn't appear to be redundant with any of the terms of the right-hand side. So I don't think you need any t e^t or t e^(2 t) terms.
See if you agree with the following:
The complementary solution is a linear combination of sin(2 t) and cos(2 t).
2 sin(t) will result from a sum of multiples of sin(t) and cos(t)
cosh(t) = (e^t + e(^-t)) / 2, and will result from a sum of multiples of e^t and e^(-t)
cosh^2(t) = (e^(2 t) + 2 + e^(-2 t)) / 2 and will result from a sum of multiples of e^(2 t), e^(-2 t) and 1.
*@
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Given Solution:
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Self-critique (if necessary):
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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:
Trail form for our Particular solution is
y_p1(t) = [ (At + B)]
Plugging into our equation [t(At + B)]
y '' + alpha y ' + beta y = t + sin(t)
((At + B))’’ + (At + B)’ + (At + B)= t
() + (alpha A) + (A beta t + B beta) = t
Now setting coefficients equal,
A = 1
A + B = 0
A= 1, B= -1, so that Particular solution is
y_p1(t) = t-1
Trail form for our Particular solution is
y_p2(t) = Acos(t) + Bsin(t)
Plugging into our equation [Acos(t) + Bsin(t)]
y '' + alpha y ' + beta y = t + sin(t)
(Acos(t) + Bsin(t))’’ + (Acos(t) + Bsin(t))’ + (Acos(t) + Bsin(t)) = sin(t)
-Acos(t) - Bsin(t) + -Asin(t) + Bcos(t) + Acos(t) + Bsin(t) = sin(t)
Bcos(t) - Asin(t) = sin(t)
Now setting coefficients equal,
A = 1
B = 0
A= 1, B= 0, so that Particular solution is
y_p2(t) = cos(t)
General solution is given by y(t) = y_c(t) + y_p1(t) + y_p2, so we have
y(t) = c_1 cos(t) + c_2 sin(t) + t - 1 + cos(t)
???I’m confused??????
@& You can determine alpha and beta right off the bat from the given complementary solution, which can only result from the equation r^2 + 1 = 0:
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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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 will be a complex number.
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’’ - y = e^(`i * 2 t), plugging in A e^(i * 2 t)
A(2`i)^2 e^(i * 2 t) - A e^(i * 2 t) = e^(i * 2 t)
A(2`i)^2 - A = 1
A( (2`i)^2 - 1) = 1
A = 1/( (2`i)^2 - 1) = -1/5????
Not so sure about second part, are we to turn e^(`i*2t) into e^(`i*t)^2?????
@&
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.
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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:
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Self-critique (if necessary):
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Self-critique rating:
@& Very good work. See my notes.*@