代做MATH-UA.0262: Ordinary Differential Equations调试R语言程序

MATH-UA.0262: Ordinary Di↵erential Equations

4. Homework 4

Exercise 4.1 (2.1.3). Show that the operator L defined by

is linear; that is, L[cy] = cL[y] and L[y1 + y2] = L[y1] + L[y2].

Exercise 4.2 (2.1.5).

1. Show that y1(t) = pt and y2(t)=1/t are solutions of the di↵erential equation

2t2y'' + 3ty' − y = 0                                     (3)

on the interval 0

2. Compute W[y1, y2](t). What happens as t approaches zero?

3. Show that y1(t) and y2(t) form. a fundamental set of solutions of (3) on the interval 0

4. Solve the initial-value problem

Exercise 4.3 (2.1.7). Compute the Wronskian of the following pairs of functions.

1. sin(at), cos(bt)

2. sin2(t), 1 − cos(2t)

3. eat, ebt

4. eat, teat

5. t, tln(t)

6. eat sin(bt), eat cos(bt).

Exercise 4.4 (2.1.9).

1. Let y1(t) and y2(t) be solutions of y00 + p(t)y0 + q(t)y = 0 on the interval α < t < β, with y1(t0)=1, y1 0(t0)=0, y2(t0)=0 and y2 0(t0)=1. Show that y1(t) and y2(t) form. a fundamental set of solutions on the interval α

2. Show that y(t) = y0y1(t) + y0 0 y2(t) is the solution of the initial value problem

5. Homework 5

Exercise 5.1 (2.2.5). Solve the following initial value problem:

Exercise 5.2 (2.2.9). Let y(t) be the solution of the initial value problem

For what values of V does y(t) remain nonnegative for all t ≥ 0?

Exercise 5.3 (2.2.1.3). Find the general solution of

y'' + 2y' + 3y = 0.

Exercise 5.4 (2.2.1.5). Solve the initial value problem

Exercise 5.5 (2.2.2.3). Solve the initial value problem

Exercise 5.6 (2.2.2.9). Here is an alternate and very elegant way of finding a second solution y2(t) of

ay'' + by0 + cy = 0

when ar2 + br + c = 0 has a double root.

1. Assume that b2 = 4ac. Show that

L[ert] = a(ert)'' + b(ert)' + cert = a(r − r1) 2ert

for r1 = −2a/b.

2. Show that

3. Conclude that L[ter1t]=0. Hence, y2(t) = ter1t is a second solution of the above constant coefficient equation in the double root case.

Exercise 5.7 (2.3.1). Three solutions of a certain second-order nonhomogeneous linear equation are

ψ1(t) = t 2, 2(t) = t 2 + e2t

and

ψ3(t)=1+ t2 + 2e2t.

Find the general solution of this equation.

Exercise 5.8 (2.4.1). Find the general solution of

Exercise 5.9 (2.4.5). Solve the initial value problem

Exercise 5.10 (2.5.1). Find a particular solution to

y'' + 3y = t3 − 1.

Exercise 5.11 (2.5.3). Find a particular solution to

y'' − y = t2et.

Exercise 5.12 (2.5.9). Find a particular solution to

y'' − 2y' + 5y = 2 cos2 t.

Exercise 5.13 (2.5.13). Find a particular solution to

y'' − 3y' + 2y = et + e2t .

6. Homework 6

In all of my solutions for §2.6 problems, I’ll elide the units in most of the computations since all problems give standard SI units. We also always choose the downward direction as the positive direction for vertical oscillation, and the rightward direction as positive for horizontal oscillation.

Exercise 6.1 (2.6.1). It is found experimentally that a 1 kg mass stretches a spring 49/320 m. If the mass is pulled down an additional 1/4 m and released, find the amplitude, period and frequency of the resulting motion, neglecting air resistance (use g = 9.8 m/s2).

Exercise 6.2 (2.6.5). A small object of mass 1 kg is attached to a spring with spring-constant 1 N/m and is immersed in a viscous medium with damping constant 2 N · s/m. At time t = 0, the mass is lowered 1/4 m and given an initial velocity of 1 m/s in the upward direction. Show that the mass will overshoot its equilibrium position once, and then creep back to equilibrium.

Exercise 6.3 (2.6.11). A 1 kg mass is attached to a spring with spring constant k = 4 N/m, and hangs in equilibrium. An external force F(t) = (1 +t+ sin(2t) N is applied to the mass beginning at time t = 0. If the spring is stretched a length (1/2+⇡/4) m or more from its equilibrium position, then it will break. Assuming no damping present, find the time at which the spring breaks.

Exercise 6.4 (2.6.13). Determine a particular solution (t) of my'' + cy' + ky = F0 cos(ωt), of the form. ψ(t) = A cos(ωt − φ). Show that the amplitude A is a maximum when ω2 = ω0 2 − 2/1 (c/m)2. This value of ω is called the resonant frequency of the system. What happens when ω0 2 < 2/1 (c/m)2?

Exercise 6.5 (2.8.1). Find the general solution of

{y'' + ty' + y = 0.

Exercise 6.6 (2.8.5). Solve the following initial-value problem:

Exercise 6.7 (2.8.9). The equation y00 − 2ty0 + λy = 0, λ constant, is known as the Hermite di↵erential equation, and it appears in many areas of mathematics and physics.

• Find 2 linearly independent solutions of the Hermite equation.

• Show that the Hermite equation has a polynomial solution of degree n if λ = 2n. This polynomial, when properly normalized; that is, when multiplied by a suitable constant, is known as the Hermite polynomial Hn(t).

7. Homework 7

Exercise 7.1 (2.8.1.7). Find the general solution of

t2y'' + ty' + y = 0.

Exercise 7.2 (2.8.1.9). Solve the initial-value problem

on the interval 0

Exercise 7.3 (2.8.2.1). Determine whether or not t = 0 is a regular singular point of the ODE

t(t − 2)2y'' + ty' + y = 0.

Exercise 7.4 (2.8.2.5). Determine whether or not t = −1 is a regular singular point of the ODE

Exercise 7.5 (2.8.2.7). Find the general solution of

2t2y'' + 3ty' − (1 + t)y = 0.

Exercise 7.6 (2.8.2.19). Consider the equation

t 2y'' + (t 2 − 3t)y' + 3y = 0.                 (4)

1. Show that r = 1 and r = 3 are the two roots of the indicial equation of (4).

2. Find a power series solution of (4) of the form.

3. Show that y1(t) = t 3e−t .

4. Show that (4) has no solution of the form.

5. Find a second solution of (4) using the method of reduction of order. Leave your answer in integral form.





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