Math 3d Winter Term 2002

Final Examination - Solutions

1. Compute the solution of the following system

\begin{cases}ẋ_{1} = −2x_{1} + x_{2} + 2t , & x_{1}(0) = 1 \\ ẋ_{2} = x_{1} − 2x_{2} − 2t , & x_{2}(0) = 1 . \end{cases}

Solution. The solution of the equation is given by the variation of parameters formula

X(t) = e^{tA}X_{0} + ∫_{0}^{t} e^{(t−τ)A}G(τ) dτ , t > 0 ,

where

A = [\begin{matrix}−2 & 1 \\ 1 & −2 \end{matrix}] , X_{0} = [\begin{matrix}1 \\ 1 \end{matrix}] , G(t) = [\begin{matrix}2t \\ −2t \end{matrix}] = 2t [\begin{matrix}1 \\ −1 \end{matrix}] .

In order to compute the exponential of the matrix A we determine its eigenvalues and eigenvectors. The eigenvalues λ are determined through

|\begin{matrix}−2 − λ & 1 \\ 1 & −2 − λ \end{matrix}| = (λ + 2)^{2} − 1 = 0

which gives λ_{1} = −1 and λ_{2} = −3. The associated eigenvectors X_{λ} are then given as solutions of

[\begin{matrix}−1 & 1 \\ 1 & −1 \end{matrix}] X_{−1} = [\begin{matrix}0 \\ 0 \end{matrix}] , [\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}] X_{−3} = [\begin{matrix}0 \\ 0 \end{matrix}] .

We therefore obtain

X_{−1} = [\begin{matrix}1 \\ 1 \end{matrix}] , X_{−3} = [\begin{matrix}1 \\ −1 \end{matrix}] .

The solution formula becomes

X(t) = e^{t[\begin{matrix}−2 & 1 \\ 1 & −2 \end{matrix}]} [\begin{matrix}1 \\ 1 \end{matrix}] + ∫_{0}^{t} 2τ e^{(t−τ)[\begin{matrix}−2 & 1 \\ 1 & −2 \end{matrix}]} [\begin{matrix}1 \\ −1 \end{matrix}] dτ =

e^{−t} [\begin{matrix}1 \\ 1 \end{matrix}] + ∫_{0}^{t} 2τ e^{−3(t−τ)} [\begin{matrix}1 \\ −1 \end{matrix}] dτ = [\begin{matrix}e^{−t} + \frac{2}{9}e^{−3t} + \frac{2}{3}t − \frac{2}{9} \\ e^{−t} − \frac{2}{9}e^{−3t} − \frac{2}{3}t + \frac{2}{9} \end{matrix}] , t > 0 .

2. a. Compute the location t_{0} > 0 where y(t_{0}) = 0 for the function y which solves

y′ = −\frac{t}{y^{2}} , y(0) = y_{0} > 0 .

b. Compute y_{∞} = lim_{t→∞} y(t) for the solution of

y′ = (1 + y)^{2} e^{−t} , y(0) = y_{0} > 0 .

Solution. a. The solution can be computed by separating the variables to obtain

∫_{0}^{t} y′(τ)y(τ)^{2} dτ = \frac{1}{3}y^{3}(t) − \frac{1}{3}y^{3}(0) = − ∫_{0}^{t} τ dτ = −\frac{1}{2}t^{2} ,

which gives y(t) = (y_{0}^{3} − \frac{3}{2}t^{2})^{1/3}. The zero of interest is therefore t_{0} = \sqrt{\frac{2}{3}y_{0}^{3}} .

b. The solution can again be computed by separation

− ∫_{0}^{t} \frac{y′(τ)}{(1 + y(τ))^{2}} dτ = \frac{1}{1 + y(t)} − \frac{1}{1 + y(0)} = ∫_{0}^{t} e^{−τ} dτ = 1 − e^{−t} ,

to obtain y(t) = _{\frac{1}{\frac{1}{1+y_{0}} +1−e^{−t}}} − 1 and finally

lim_{t→∞} y(t) = \frac{1}{\frac{1}{1+y_{0}} + 1} − 1 = −\frac{1}{2 + y_{0}} .

3. a. Solve

\begin{cases}y′′ − 2y′ + y = 1 + e^{−t} , \\ y(0) = 0 , y′(0) = 1. \end{cases}

b. Find the solution of

\begin{cases}4y′′ + 4y′ + 5y = 0 , \\ y(0) = 1 , y′(0) = 0. \end{cases}

Solution a. The characteristic equation λ^{2} − 2λ + 1 has the double root λ = 1. The general solution of the homogeneous equation is therefore given by

y_{h}(t) = c_{1}e^{t} + c_{2}t e^{t} .

A particular solution y_{i} of the inhomogeneous equation can be looked for in the form

y_{i}(t) = A + Be^{−t} .

Plugging this Ansatz into the equation gives

Be^{−t} − 2(−Be^{−t}) + A + Be^{−t} = A + 4Be^{−t} = 1 + e^{−t} ,

from which follows A = 1 and B = 1/4. Finally the constants c_{1} and c_{2} have to be determined imposing the initial condition

y_{h}(0) + y_{i}(0) = c_{1} + 1 + 1/4 = 0 , y′_{h}(0) + y′_{i}(0) = c_{1} + c_{2} − 1/4 = 1 .

This gives c_{1} = −5/4 and 5/2.

b. The characteristic equation 4λ^{2} + 4λ + 5 can be solved to obtain λ_{1,2} = −\frac{1}{2} ± i which leads to the general solution

y(t) = c_{1}e^{−t/2} cos(t) + c_{2}e^{−t/2} sin(t) .

The initial condition gives

y(0) = c_{1} = 1 , y′(0) = −c_{1}/2 + c_{2} = 0 and finally c_{1} = 1 , c_{2} = 1/2 .

4. Let 0 < ε < 1 and solve

\begin{cases}y′′ + 2εy′ + y = cos(t) , \\ y(0) = 0 , y′(0) = 1. \end{cases}

What is the amplitude of the solution for large time?

Solution. The characteristic equation has solutions λ_{1,2} = −ε ± i\sqrt{1 − ε^{2}} and therefore

y_{h}(t) = c_{1}e^{−εt} cos(t\sqrt{1 − ε^{2}}) + c_{2}e^{−εt} sin(t\sqrt{1 − ε^{2}}) .

We can seek a solution y_{i} to the inhomogeneous equation in the form

y_{i}(t) = A cos(t) + B sin(t) .

Plugging into the equation

− A cos(t) − sin(t)B + 2ε(−A sin(t) + B cos(t)) + A cos(t) + B sin(t) =

(−A + 2εB + A) cos(t) + (−B − 2εA + B) sin(t) = 2εB cos(t) − 2εA sin(t) = cos(t)

we see that A = 0 and B = 1/2ε. The initial condition then gives

y(0) = y_{h}(0) + y_{i}(0) = c_{1} = 0 , y′(0) = c_{2}\sqrt{1 − ε^{2}} + 1/2ε = 1 .

and the solution is

y(t) = \frac{2ε − 1}{2ε\sqrt{1 − ε^{2}}} e^{−εt} sin(t\sqrt{1 − ε^{2}}) + sin(t)/2ε .

Its asymptotic amplitude is therefore 1/2ε.

5. Solve the following linear system

\begin{cases}ẋ_{1} = −2x_{1} + x_{2} , & x_{1}(0) = 0 \\ ẋ_{2} = −2x_{2} + x_{3} , & x_{2}(0) = 1 \\ ẋ_{3} = −2x_{3} , & x_{3}(0) = 1 \end{cases}

Solution. We need to compute the exponential of the matrix M = [\begin{matrix}−2 & 1 & 0 \\ 0 & −2 & 1 \\ 0 & 0 & −2 \end{matrix}]. Consider the matrix

A(λ) = [\begin{matrix}λ & 1 & 0 \\ 0 & λ & 1 \\ 0 & 0 & λ \end{matrix}] , A(−2) = M ,

and recall that

e^{tA(λ)} = ∑_{k=0}^{∞} (tA(λ))^{k}/k! .

Computing the powers of A(λ) we see that

A(λ)^{2} = [\begin{matrix}λ^{2} & 2λ & 1 \\ 0 & λ^{2} & 2λ \\ 0 & 0 & λ^{2} \end{matrix}] , A(λ)^{3} = [\begin{matrix}λ^{3} & 3λ^{2} & 3λ \\ 0 & λ^{3} & 3λ^{2} \\ 0 & 0 & λ^{3} \end{matrix}] , A(λ)^{4} = [\begin{matrix}λ^{4} & 4λ^{3} & 6λ^{2} \\ 0 & λ^{4} & 4λ^{3} \\ 0 & 0 & λ^{4} \end{matrix}]

Plugging into the series we obtain

1 + tλ + t^{2}λ^{2}/2 + ⋯ + t^{k}λ^{k}/k! + ⋯ = e^{λt}

on the diagonal. On the second and third upper diagonals we have

0 + t + t^{2}2λ/2 + t^{3}3λ^{2}/3! + ⋯ + t^{k}kλ^{k−1}/k! = t e^{λt} and 0 + 0 + t^{2}/2 + t^{3}3λ/3! + t^{4}6λ^{2}/4! + ⋯ + t^{k}(k − 1)kλ^{k−2}/k! + ⋯ = t^{2} e^{λt}/2 , respectively. We finally obtain

X(t) = e^{tA(2)} [\begin{matrix}0 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix}e^{2t} & t e^{2t} & t^{2} e^{2t}/2 \\ 0 & e^{2t} & t e^{2t} \\ 0 & 0 & e^{2t} \end{matrix}] [\begin{matrix}0 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix}t e^{2t} + t^{2} e^{2t}/2 \\ e^{2t} + t e^{2t} \\ e^{2t} \end{matrix}]

6. Solve

\begin{cases}(1 + t^{2})y′′ + y = 0 , \\ y(0) = 0 , y′(0) = 1. \end{cases}

expanding the solution in a power series about t = 0 and determine its radius of convergence. Compute only the first five coefficients of the expansion.

Solution. We look for the solution in the form ∑_{k=0}^{∞} a_{k}t^{k}. The radius of convergence of the series is given by the distance between t = 0 and the closest zero of t^{2} + 1. Since the zeros are ±i we obtain a radius of convergence ρ = 1. Plugging the series Ansatz into the equation we obtain

(1 + t^{2}) ∑_{k=2}^{∞} a_{k}k(k − 1)t^{k−2} + ∑_{k=0}^{∞} a_{k}t^{k} =

a_{0} + 2a_{2} + t[6a_{3} + a_{1}]t + ∑_{k=2}^{∞}[(k + 1)(k + 2)a_{k+2} + (k^{2} − k + 1)a_{k}]t^{k}

The initial condition entails a_{0} = 0 and a_{1} = 1. The remaining coefficients can be obtained via the recurrence relation

a_{k+2} = −\frac{k^{2} − k + 1}{(k + 1)(k + 2)}a_{k} , k = 0, 1, 2, …

to give a_{2} = 0, a_{3} = −1/6, a_{4} = 0.

7. Consider the following equations

(x^{2} − 1)y′′ + sin((x − 1)π/4)y′ + (x^{2} − 3x + 2)y = 0 ,
(x^{3} − 1)y′′ + 2xy′ − y = 0 ,
xy′′ + sin(\frac{1}{x})y′ + 2y = 0 .

a. Determine their singular points and whether they are regular?

b. For the first equation determine the exponents at the singularity.

c. What can you say about the radius of convergence of a series solution of the second equation about x = 1.

Solution a. The point x = −1 is a regular singular point for the first equation, x = 1 a regular singular point of the second whereas x = 0 is not a regular singular point for the last equation. b. The exponents at the singularity for the first equation are the solutions of the inditial equation r(r − 1) + p_{0}r + q_{0}r^{2} = 0 where

p_{0} = lim_{x→−1}(x + 1)\frac{sin((x − 1)π/4)}{(x + 1)(x − 1)} = 1/2 , q_{0} = lim_{x→−1}(x + 1)^{2} \frac{(x − 2)(x − 1)}{(x + 1)(x − 1)} = 0 .

We therefore obtain r_{1,2} = 0, 1/2. c. The other solutions of x^{3} − 1 = (x^{2} + x + 1)(x − 1) = 0 are given by

x_{1,2} = \frac{1}{2}(−1 ± \sqrt{1 − 4}) = −\frac{1}{2} ± i\frac{\sqrt{3}}{2} .

The distance to the closest zero from x = 1 is given by \sqrt{3}. The radius of convergence is therefore at least \sqrt{3}.

8. A bungee jumper jumps off a 180m high bridge attached to a 125m long cord. His body mass is m = 80kg and the cord behaves like a spring with constant k = 160kg/sec^{2}. How close to the ground does the jumper come? Neglect air resistance and assume g = 10m/sec^{2}.

Solution. The equation which determines the displacement from the equilibrium is given by

m y′′ + k y = 80 y′′ + 160 y = 0 .

To determine the initial conditions we need to find the speed of the jumper when the cord is fully elongated. Using g = 10 m/sec^{2} we see that the jumper has fallen 125 m after t = 5 sec as can computed from

\frac{1}{2}10t^{2} = 125 .

This gives a velocity

y′(0) = 50 [m/sec] .

Since the spring constant is k = 160 kg/sec^{2} and g = 10 m/sec^{2} we see that the elongation of the cord at equilibrium is given by \frac{80·10}{160} \frac{kg m/sec^{2}}{kg/sec^{2}} = 5 m. We finally obtain the initial condition

y(0) = −5 [m] .

Finally we obtain the equation

y′′ + 2y = 0 , y(0) = −5 , y′(0) = 50 .

Since the general solution is given by

c_{1} cos(t\sqrt{2}) + c_{2} sin(t\sqrt{2})

we obtain

c_{1} = −5 and c_{2}\sqrt{2} = 50 ,

which gives an amplitude of R = \sqrt{25 + 2 ∗ 25^{2}} = 5\sqrt{51} ≈ 35. The jumper therefore falls a total of about 165 m.

9. Solve

\begin{cases}ẋ_{1} = −x_{1} − 2x_{2} , & x_{1}(0) = 1 \\ ẋ_{2} = 2x_{1} − x_{2} , & x_{2}(0) = 1 . \end{cases}

Solution. The characteristic equation of A = [\begin{matrix}−1 & −2 \\ 2 & −1 \end{matrix}] is given by (−1 − λ)^{2} + 4 = 0 which gives the eigenvalues

λ_{1,2} = −1 ± 2i .

The associated eigenvectors are given by

X_{−1+2i} = [\begin{matrix}i \\ 1 \end{matrix}] , X_{−1−2i} = [\begin{matrix}i \\ −1 \end{matrix}] .

We can use them to obtain

[\begin{matrix}1 \\ 0 \end{matrix}] = \frac{1}{2i}(X_{−1+2i} + X_{−1−2i}) and [\begin{matrix}0 \\ 1 \end{matrix}] = \frac{1}{2}(X_{−1+2i} − X_{−1−2i}) .

Since the solution is given by

e^{tA} [\begin{matrix}1 \\ 1 \end{matrix}] = e^{tA}([\begin{matrix}1 \\ 0 \end{matrix}] + [\begin{matrix}0 \\ 1 \end{matrix}])

we compute

e^{tA} [\begin{matrix}1 \\ 0 \end{matrix}] = \frac{1}{2i}e^{tA}(X_{−1+2i} + X_{−1−2i}) = \frac{1}{2i}(e^{−t+2it}X_{−1+2i} + e^{−t−2it}X_{−1−2i}) = e^{−t} [\begin{matrix}cos(2t) \\ sin(2t) \end{matrix}]

e^{tA} [\begin{matrix}1 \\ 0 \end{matrix}] = \frac{1}{2}e^{tA}(X_{−1+2i} − X_{−1−2i}) = \frac{1}{2}(e^{−t+2it}X_{−1+2i} − e^{−t−2it}X_{−1−2i}) = e^{−t} [\begin{matrix}− sin(2t) \\ cos(2t) \end{matrix}]

which gives the solution

e^{tA} [\begin{matrix}1 \\ 1 \end{matrix}] = e^{−t} [\begin{matrix}cos(2t) − sin(2t) \\ cos(2t) + sin(2t) \end{matrix}]

10. Determine the indicial equation and the corresponding exponents at the singularity for the regular singular points of

x^{2}y′′ + x^{2}y′ − y = 0
x sinh(x)y′′ + xy′ − y = 0
tanh(x)y′′ − y′ = 0

Solution. The point x = 0 is the only regular singular point for each equation. The indicial equations and the corresponding exponents are given by

r(r − 1) − 1 = 0 , r_{1,2} = \frac{1}{2} ± \frac{\sqrt{5}}{2} ,

r(r − 1) + r − 1 = 0 , r_{1,2} = ±1 , r(r − 1) + r = 0 , r_{1,2} = 0 .