Download MATH 23 Homework #3 (Part B) Solutions

MATH 23 Homework #3 (Part B) Solutions
October 7, 2016
Problem 3.3.15 Find the general solution of the given differential equation
y 00 + y 0 + 1.25y = 0
Solution The characteristic equation is r2 + r + 1.25 = 0 , with roots r =
−(1/2) ± i. Hence the general solution is y = c1 e−t/2 cos t + c2 e−t/2 sin t .
Problem 3.3.19 Find the solution of the given initial value problem. Sketch
the graph of the solution and describes its behaviour as t increases.
Solution The characteristic equation is r2 − 2r + 5 = 0, with roots r = 1 ± 2i.
Hence the general solution is y = c1 et cos 2t + c2 et sin 2t.
Based on the initial condition y(π/2) = 0, we require that c1 = 0. It follows
that y = c2 et sin 2t, and so the first derivative is y 0 = c2 et sin 2t + 2c2 et cos 2t.
In order to satisfy the condition y 0 (π/2) = 2, we find that −2eπ/2 c2 = 2.
Hence we have c2 = −e−π/2 . Therefore the specific solution is y(t) =
−et−π/2 sin 2t. The solution oscillates with an exponentially growing amplitude.
Problem 3.3.26 Consider the IVP
y 00 + 2ay 0 + (a2 + 1)y = 0,
y(0) = 1,
y 0 (0) = 0
(a) Find the solution y(t) of this problem
(b) For a = 1 find the smallest T such that |y(t)| < 0.1 for t > T
(c) Repeat part (b) for a = 1/4, 1/2, 2
(d) Using the results of parts (b) and (c), plot T versus a and describe the
relation between T and a.
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Solution
(a) The characteristic equation is r2 +2ar+(a2 +1) = 0, with roots r = −a±i.
Hence the general solution is y(t) = c1 e−at cos t + c2 e−at sin t .
Based on the initial conditions, we find that c = 1 and c = a . Therefore
specific solution is given by y(t) = e−at cos t + ae−at sin t =
√ the −at
1 + a2 e cos(t − φ), in which φ = arctan(a).
√
(b) For estimation,
note
that
|y(t)|
≤
1 + a2 e−at . Now consider the √
in√
−at
2
2
equality 1 + a e
≤ 1/10
√ . The inequality holds for t ≥ (1/a) ln(10 1 + a ).
Therefore T ≥ (1/a) ln(10 1 + a2 ). Setting a = 1,the numerical value is
T ≈ 1.8763.
(c) Similarly, T1/4 ≈ 7.4284, T1/2 ≈ 4.3003, T2 ≈ 1.5116.
√
(d) Note that the estimates Ta approach the graph of (1/a) ln(10 1 + a2 ) as
a gets large.
Problem 3.4.12 Solve the given IVP. Sketch the graph of the solution and
describe its behaviour for increasing t.
y 00 − 6y 0 + 9y = 0,
y(0) = 0,
y 0 (0) = 2
Solution Characteristic equation is r2 − 6r + 9 = 0. The roots are r1,2 = 3, 3.
So the general solution is
y(t) = (c1 + c2 te3t ).
Using the initial conditions, we get c1 = 0, c2 = 2.
So the solution is y(t) = 2te3t . As t → ∞, y → ∞.
Problem 3.4.15 Consider the IVP
4y 00 + 12y 0 + 9y = 0,
y(0) = 1,
2
y 0 (0) = −4
(a) Solve the IVP and plot its solution for 0 ≤ t ≤ 5.
(b) Determine where the solution has value zero
(c) Determine the coordinates (t0 , y0 ) of the minimum point.
(d) Change the second initial condition to y 0 (0) = b and find the solution as
a function of b. Then find the critical value of b that separates solutions
that always remain positive from those that eventually become negative.
Solution
(a) The characteristic equation is 4r2 + 12r + 9 = 0 , with double root
r = −3/2. The general solution is y(t) = c1 e−3t/2 + c2 te−3t/2 . Invoking the first initial condition, it follows that c1 = 1 . Now y 0 (t) =
(−3/2 + c2 )e−3t/2 − (3/2)c2 te−3t/2 . The second initial condition requires
that −3/2 + c2 = −4 , or c2 = −5/2 . Hence the specific solution is
y(t) = e−3t/2 − (5/2)te−3t/2 .
(b) The solution crosses the x-axis at t = 2/5.
(c) The solution has a minimum at the point (16/15, −5e−8/5 /3).
(d) Given that y 0 (0) = b, we have −3/2 + c2 = b, or c2 = b + 3/2. Hence
the solution is y(t) = e−3t/2 + (b + 3/2)te−3t/2 . Since the second term
dominates, the long-term solution depends on the sign of the coefficient
b + 3/2. The critical value is b = −3/2.
Problem 3.4.30 Use the method of reduction of order to find a second
solution to the given differential equation
x2 y 00 + xy 0 + (x2 − 0.25)y = 0,
x > 0;
y1 (x) = x−1/2 sin x
Solution Suppose y(x) = v(x)y1 (x) = v(x)x−1/2 sin x. Taking the derivatives
we get,
y 0 (x) = v 0 x−1/2 sin x − (1/2)vx−3/2 sin x + vx−1/2 cos x,
y 00 (x) = v 00 x−1/2 sin x − v 0 x−3/2 sin x + v 0 x−1/2 cos x − vx−3/2 cos x + (3/4)vx−5/2 sin x − vx−1/2 sin x.
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After substituting y 00 , y 0 and y into the DE, we get
v 00 + 2v 0 cot x = 0
Suppose v 0 = w and v 00 = w0 , we get w0 + 2w cot x = 0. On solving, we
get w = c1 (sin x)−2 . On integrating, we get v(x) = −c1 cot x + c2 . Thus,
y(x) = c3 x−1/2 cos x (after dropping the term that appears in y1 (x)).
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