Download Differential Equations Math 225 Exam 2 (Sample) October, 2005

Differential Equations Math 225
Exam 2 (Sample)
October, 2005
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 y'= 2y + 3z + 2
1. (30 pts) Solve the system of equations: 
 z'= y − 2z + t
Method 1 – Elimination:
This is equivalent to the system (written in terms of linear differential operators):
(D-2)[y] + (-3)[z] = 2
(*)
€
(-1)[y] + (D+2)[z] = t
(**)
Apply the appropriate operators so the y terms of both are the same, so we can
subtract to get an equation involving only x:
(-1)(D-2)[y] + (-1)(-3)[z] = (-1)[2] = -2
(D-2)(-1)[y] + (D-2)(D+2)[z] = (D-2)[t] = 1-2t
Subtracting we get ((3) – (D-2)(D+2))[z] = -2 – (1-2t), so –z” + 7z = 2t-3
Solving this equation in the usual manner we proceed as follows:
Char polynomial is -r2 + 7 = 0, which has roots r = {±√7}
Thus the homogeneous solution is z H ( t ) = Ae 7t + Be− 7t
Using the method of undetermined coefficients we get zP(t) = (2t-3)/7
(We can actually do this by inspection in this case, as the z” term will be zero)
Thus z(t) = zH(t) + zP(t) = Ae√7t + B e-√7t + (2t-3)/7
We can now finish by plugging
€ back into equation (**), getting an equation in y:
(-1)[y] + (D+2)[z] = t
-y + (A√7e√7t – B√7e-√7t + 2/7) + 2(Ae√7t + B e-√7t + (2t-3)/7) = t
Answer:
y = A(2+√7)e√7t + B(2-√7)e-√7t + (-3t-4)/7
z = Ae√7t + B e-√7t + (2t-3)/7
(Note: This method can introduce extraneous constants, but when we plugged back into eqn (**) to
solve for y we also solved for these constants. Had we solved for y in the same way that we
solved for z – eliminate by subtracting “multiples” of (*) and (**) we would have gotten two
further constants – call them C and D – which we would have had to solve for in terms of A and B
in by plugging our solutions back into the two equations.)
Method 2 – Matrix Methods:
This system of equations is equivalent to the matrix equation X’ = AX + B, where
 y
2 3 
2
X =  , A = 
 and B =  
z
1 −2
t
2 + 7 
2 − 7 
The matrix A has eigenvalues {√7, -√7}. The eigenvectors 
 and 

 1 
 1 
€correspond to the€eigenvalues √7 and -√7, respectively. (In other words,
2 + 7 
2 − 7 
2 3  2 + 7 
2 3  2 − 7 
 = 7
 and 
 = − 7
 .) Using



1 
1 −2 1 
1 −2 1€ 
 1 
 €
these eigenvectors we can simplify our problem by diagonalizing A:
We create a matrix T from these eigenvectors, knowing that the matrix T-1AT will be
diagonal, with eigenvalues on the diagonal. Think of this as a change of variables,
from {y,z}, to {T-1(y,z)}. We
€ convert our original equation X’ = AX + B by
€
€
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Exam 2 Sample
multiplying by T-1 to get the equation T-1X’ = T-1AX + T-1B. With a little
manipulation we get the equation T-1X’ = T-1ATT-1X + T-1B. By defining some new
names: U:= T-1X, the vector of (changed) variables, and D:= T-1AT, the diagonalized
matrix, and C:= T-1B, we can rewrite this as the equation: U’ = DU + C.
 2 + 7 2 − 7 −12 3  2 + 7 2 − 7  2 + 7   7
0 
−1
D = T AT = 
 
=
=


1  1 −2 1
1   1   0 − 7
 1
 2 + 7 2 − 7 −1 y 
v 
−1
U = =T X =
  
1   z
w
 1
 7

7
 2 + 7 2 − 7 −1 2  7 1+ 2 −1 t 
−1
C =T B =
  =
1   t   7 −1+ 7 + 1 t 
 1
2
7

This gives us the two first order equations:
€
( (
))
( ( ))
€
v'= 7v +
7
7
(1+ ( −1)t)
(−1+ ( + 1)t)
7
2
€
w'= 7w +
€
Solvable by the usual first order methods, after which we can convert back to y and z
as {y, z} = {T(v,w)}.
€ 2.
7
7
7
2
(30 pts) Given the equation y” + 2y’ + 5y = 0 :
a) (15 pts) Find the general solution.
The characteristic polynomial r2 + 2r + 5 = 0 has roots r = {-1 ± 2i}
The general (which in this case is the homogeneous) solution is:
Answer: y(t) = a1 e(-1+2i)t + a1 e(-1-2i)t which can be rewritten as either
y(t) = e-t(d1 sin(2t) + d2 cos(2t)) or
y(t) = Ae-tsin(2t + φ)
b) (10 pts) Solve the initial value problem y(π) = 1, y’(π) = 2.
y(t) = e-t(d1 sin(2t) + d2 cos(2t))
y’(t) = -e-t(d1 sin(2t) + d2 cos(2t)) + y(t) + e-t(2d1 cos(2t) - 2d2 sin(2t))
= e-t((-d1 -2d2 )sin(2t) + (-d2 +2d1 )cos(2t))
If t=π we have sin(2t) = sin(2π) = 0 and cos(2t) = cos(2π) = 1
So 1 = y(π) = e-π(d1 (0) + d2 (1)) = d2 e-π and thus d2 = eπ
And 2 = y’(π) = e-π((-d1 -2d2 )(0) + (-d2 +2d1 )(1)) = e-π(-d2 +2d1 ) = e-π(- eπ+2d1 )
2 = -1 + 2d1 e-π and thus d1 = 3eπ/2
Answer: y(t) = e-t((3eπ/2)sin(2t) + (eπ)cos(2t)) = eπ-t((3/2)sin(2t) + cos(2t))
c) (5 pts) Characterize the solution (oscillatory – if so what frequency,
exponentially dying or growing, etc)
The solution is oscillatory in an exponentially decreasing envelope. The
oscillations have period π.
3. (30 pts) Solve 2y” + 8y = sin(2t):
We first solve the homogeneous equation 2y” + 8y = 0
2/4
Exam 2 Sample
The characteristic polynomial is 2r2 + 8 = 0, which has roots r = {±2i}
So yH(t) = a1 e2it + a2 e-2it or equivalently
yH(t) = d1 sin(2t) + d2 cos(2t)
a) (15 pts) By undetermined coefficients.
Normally, a driving term of sin(2t) suggests a particular solution of form
Asin(2t) + Bcos(2t). In this case, though, as the driving term is of the same
form of part of the homogeneous solution, we will look for a particular solution
of form At sin(2t) + Bt cos(2t). (text, pg 184)
(This can also be made plain by noting that the equation has the form 2(D+2)(D-2)[y] =
sin(2t), and as (D+2)(D-2) annihilates sin(2t) we can rewrite this as a homogeneous equation
by operating on both sides with (D+2)(D-2), so 2(D+2)2(D+2)2[y] = (D+2)(D-2)[sin(2t)] = 0.
The usual reduction of order argument gives us the form of the solutions.)
Assume that yp (t) = c1 t sin(2t) + c2 t cos(2t)
yp ’(t) = c1 sin(2t) + 2c1 t cos(2t) + c2 cos(2t) - 2c2 t sin(2t)
yp ”(t) = 4c1 cos(2t) - 4c1 t sin(2t) - 4c2 sin(2t) - 4c2 t cos(2t)
Plugging back into the original equation:
2(4c1 cos(2t)-4c1 tsin(2t)-4c2 sin(2t)-4c2 tcos(2t))+8(c1 tsin(2t)+c2 tcos(2t))=sin(2t)
As {sin(2t), cos(2t), t sin(2t), t cos(2t)} are linearly independent functions, we
can divide this into four separate relations, looking at the coefficients of each of
these functions:
sin(2t): -8c2 = 1, so c2 = -1/8
cos(2t): 8 c1 = 0, so c1 = 0
t sin(2t): -8c1 + 8c1 = 0, which gives no information
t cos(2t): -8c2 + 8c2 = 0, which gives no information
Answer: y(t) = yH(t) + yP(t) = d1 sin(2t) + d2 cos(2t) - (1/8)t cos(2t)
b) (15 pts) By variation of parameters.
Assume that the particular solution has the form:
yp (t) = u1 (t)sin(2t) + u2 (t)cos(2t)
yp ’(t) = u1 ’(t)sin(2t) + 2u1 (t)cos(2t) + u2 ’(t)cos(2t) - 2u2 (t)sin(2t)
We assume the usual constraint: u1 ’(t)sin(2t) + u2 ’(t)cos(2t) = 0, which we will
label as relation (*) for future reference.
So yp ’(t) = 2u1 (t)cos(2t) - 2u2 (t)sin(2t)
So yp ”(t) = 2u1 ’(t)cos(2t) - 4u1 (t)sin(2t) - 2u2 ’(t)sin(2t) - 4 u2 (t)cos(2t)
Plugging these back into the original equation:
2(2u1 ’(t)cos(2t) - 4u1 (t)sin(2t) - 2u2 ’(t)sin(2t) - 4 u2 (t)cos(2t)) + 8(u1 (t)sin(2t)
+ u2 (t)cos(2t)) = sin(2t)
4u1 ’(t)cos(2t) - 4u2 ’(t)sin(2t) = sin(2t), which we label as relation (**)
Multiply (*) by 4 cot(2t) and subtract it from (**)
-4u2 ’(t)cos(2t)cot(2t) - 4u2 ’(t)sin(2t) = sin(2t)
-4 u2 ’(t) = sin(2t)2
So u2 (t) = (-1/4)(t/2 – sin(4t)/8)
Plugging this back into (*) we get:
0 = u1 ’(t)sin(2t) + u2 ’(t)cos(2t) = u1 ’(t)sin(2t) + (-sin(2t)2 /4)cos(2t)
So u1 ’(t) = sin(2t)cos(2t)/4 = sin(4t)/8
u1 (t) = -cos(4t)/32
Plugging this back in to recover the particular solution:
yp (t) = u1 (t)sin(2t) + u2 (t)cos(2t) = (-cos(4t)/32)sin(2t) + ((-1/4)(t/2 –
sin(4t)/8))cos(2t) = sin(2t)/16 – (t/8)cos(2t)
Thus, adding the particular solution to the general solution we get:
y(t) = yH(t) + yP(t) = (d1 sin(2t) + d2 cos(2t)) + sin(2t)/16 + (-t/8)cos(2t)
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Exam 2 Sample
We can simplify this by defining new (still undetermined) constant coefficients
c1 = d1 + 1/16 and c2 = d2 , so we get:
Answer: y(t) = c1 sin(2t) + c2 cos(2t) + (-t/8)cos(2t)
4.
(10 pts) State an answer and if possible give a reason.
a) State whether the set of functions {x, |x|} linearly dependent over each of the
intervals [-1,0], [0,1], [-1,1].
Over the interval [0,1] the function |x| is equal to the function (1)x, so obviously
the set {x, |x|} = {x, x} is not linearly independent.
Over the interval [-1, 0] the function |x| is equal to the function (–1)x, so
obviously the set {x, |x|} = {x, (-1)x} is not linearly independent.
Over the interval [-1, 1] the two functions {x, |x|} are linearly independent. This
can be shown by noting that any relation between the two functions which
holds for x>0 has the form (a)x + (-a)|x| = 0 but any relation between them
which holds for x<0 must have the form (a)x + (a)|x| = 0. Thus, for the
relation to hold for both x<0 and x>0 we must have a=0. Thus, {x, |x|} is
linearly independent on the interval [-1, 1].
b) What is the behavior as t->∞ of the solution to the initial value problem given by
y"−( 3 + 2e−t ) y'+(2 + 3e−t + e−2t ) y = 0 and y(0) = y’(0) = 1.
€
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The solution is unbounded. More specifically, the limit as t->∞ is -∞.
We think of this equation as though it were a linear equation (it is) with constant
coefficients (it isn’t), and analyze it in the usual manner.
The characteristic polynomial is r2 – (3+2e-t)r + (2+3e-t+e-2t) = 0
which has roots r = {1+e-t, 2+e-t}, so we expect that in the interval [0, ∞] there
will be two fundamental solutions {y1 (t), y2 (t)}. One fundamental solution, call
it y1 (t), will be greater than et and less than e2t, while y2 (t) will be greater than e2t
and less than e3t – in particular, both fundamental solutions are unbounded.
(We conclude this as the real part of the two roots are both positive.) As the
initial condition is set at t=0, where the two solutions behave like y1 (t)≈e2t and
y2 (t)≈e3t, respectively, we expect that the solution will have the form y(t) = 2y1 (t)
+ 3y1 (t), and thus we expect that the solution will eventually turn negative and
have a limit as t->∞ of -∞.
Exam 2 Sample