#$&*
course MTH 279
4/9 3:23 pm
Query 19 Differential Equations*********************************************
Question: Find the general solution of the equation
y '' + y = e^t sin(t).
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Your solution:
complimentary solution
r^2+1=0
r= 0±1i
y_c = c1 cos t + c2sin t
g(t) = e^t sin t
Assumed form:
y = t^r*e^(alpha*t)(A sin beta*t + B cos beta*t
alpha = 1, beta = 1, and r=0 since y_c and y_p don't have any common terms because of the e^t
y_p= e^t(A sin t + B cos t)
y'= e^t(A sin t + B cos t) + e^t ( A cos t - B sin t)
y''= e^t(A sin t + B cos t)+ e^t ( A cos t - B sin t)+ e^t( A cos t - B sin t)+ e^t(-A sin t - B cos t)
reduced:
y''= e^t(2A cos t - 2B sin t)
sub:
e^t(2A cos t - 2B sin t) + e^t(A sin t + B cos t) = e^t sin t
2A cos t - 2B sin t + A sin t + B cos t = sin t
2 eqs 2 unknowns
2A + B = 0
A-2B = 1
A = 1/5
B = -2/5
y=c1 cos t + c2 sin t +e^t(1/5 sin t - 2/5 cos t)
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Question: Find the general solution of the equation
y '' + y ' = 6 t^2
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Your solution:
r^2 + r = 0
r = 0, r= -2
y= c1-c2e^-t
6t^2 --> assume form--> t^r(At^2+Bt+C)
here r= 1 since C is in y_c as c1
y_p = At^3+Bt^2+Ct
y'= 3At^2 + 2Bt+ C
y''= 6At+2B
sub
(6At+2B) + (3At^2+2Bt+C) = 6t^2
3A= 6
A = 2
6A+ B= 0
B = -6
2B + C= 0
C= -12
y_p = 2t^3-6t^2-12t
y= c1-c2e^-t + 2t^3-6t^2+12t
If 2 B + C = 0 and B = -6, then C = 12, not -12.
@& If correct, this would make the general solution
y(t) = y_C + y_P = A + B e^-t + 2 t^2 - 6 t + 12
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Question: Find the general solution of the equation
y '' + y ' = cos(t).
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Your solution:
r^2 + r = 0
r = 0, r= -2
y= c1-c2e^-t
cos t --> form--> t^r(e^alpha*t)( A sin beta*t + B cos beta*t)
r = 0, alpha = 0, beta = 1
y_p = A sin t + B cos t
y'= A cos t - B sin t
y''= -A sin t - B cos t
(-A sin t - B cos t)+ ( A cos t - B sin t) = sin t
the right-hand side would be cos t
-A - B = 1
A - B = 0
A = B = -1/2
the equations should have been -A - B = 1 and A - B = 0.
y_p = -1/2 sin t - 1/2 cos t
y= c1-c2e^-t - 1/2 sin t - 1/2 cos t
@&
Our complementary solution is the same as in the preceding question
y_C = A + B e^-t
Our particular solution is expected to be of the form
See if you agree:
y_P = C cos(t) + D sin(t)
so that
y_P ' = -C sin(t) + D cos(t)
and
y_P '' = -C cos(t) - D sin(t).
Substituting y_P into the equation we therefore obtain
-C cos(t) - D sin(t) - C sin(t) + D cos(t) = cos(t).
Rearranging we have
(-C + D) cos(t) + (-C - D ) sin(t) = cos(t)
so
-C + D = 1
-C - D = 0
with solutions
C = -1/2, D = 1/2.
Our particular solution is therefore
y_P = -1/2 cos(t) + 1/2 sin(t)
and our general solution is
y = y_C + y_P = A + B e^-t -1/2 cos(t) + 1/2 sin(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 '' - 2 y ' = 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t)
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Your solution:
2 e^-t cos(t) + t^2 + t e^(3 t)
y_c = e^t(c1 sin sqrt 2t + c2 cos sqrt 2t)
y_p = t(e^alpha*t)( A sin beta*t + B cos beta*t) + (Ct^2 + Dt+ E)+(e^3t)(Ft+G)
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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:
cosh (t) = 1/2(e^x + e^-x)
cosh^2(t) = 1/4(e^2x + 2 + e^-2x)
2 sin t + 1/2e^x+ 1/2 e^-x + 1/4e^2x + 1/2 + 1/4e^-2x
y_p = A sin t + B cos t + C/2(e^t + e^-t)+ D/4(e^2t+ e^-2t) + E/2
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@& The form
A sin t + B cos t + C/2(e^t + e^-t)+ D/4(e^2t+ e^-2t) + E/2
will give the same results as the form
A sin t + B cos t + C(e^t + e^-t)+ D(e^2t+ e^-2t) + E.
Extra factors in the constants do no harm, except that they can lead to errors.
However you are assuming equal coefficients of e^x and e^(-x). This might well work out since y ' is not present on the left-hand side, but in general cosh(t) will result in derivatives of both cosh(t) and sinh(t).
Similarly cosh^2(t) cannot be expected to come from a multiple of just cosh(t), so the same constant on e^(2 t) and e^(-2 t) is probably not wise.
It might be better to just go to the source functions:
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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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_C = y_C = c_1 cos(t) + c_2 sin(t)
means roots = A ± iB, where A=0 and B= 1 which means alpha = 0 and beta = 1
y''+ y = t + sin t
y_p= At + t(B sin t + C cos t)
y' = A + (B sin t + C cos t) + t(B cos t - C sin t)
y''= (B cos t - C sin t) + ( B cos t - C sin t) + t(-B sin t - C cos t)
y''= 2B cos t -2C sin t - tB sin t - tC cos t
(2B cos t -2C sin t - tB sin t - tC cos t) + (At + tB sin t + tC cos t)) = t + sin t
2B cos t - 2 C sint + At = t+ sin t
A = 1
B = 0
C = -1/2
y_p = t -1/2t cos t
y= c1 sin t + c2 cos t + t - 1/2t cos t
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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_P'= A*2i e^(i*2t)
y_P''= A*(2i)^2 e^(i*2t) = -4Ae^(2i*t)
-4Ae^(2i*t) - A e^(i * 2 t) = e^(`i * 2 t)
-4A - A = 1
This means that A = -1/5, which is not a complex number like the problem said it should be.
I can't find this problem in the book but the book says y_P will be complex valued and not A.
y= c1 e^t + c2 e^-t - 1/5 e^(2i*t)
y'= c1e^t + c2e^t-2/5ie^2it
y''= c1e^t + c2e^t +4/5 e^(2it)
(c1e^t + c2e^t +1/5 e^(2it))-(c1 e^t + c2 e^-t + 4/5*e^(2i*t)) = e^(2it)
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@& Good, but check the following expanded explanation for a generalization.
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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&#Good responses. See my notes and let me know if you have questions. &#