CS 130 Problem Set 2 22 March 2013

• Show your complete solution. • This should be done individually. • If you have consulted references for an item, cite them as footnotes to that specific item. • The problems are taken from the book “Differential Equations with Linear Algebra” by Boelkins, Goldberg and Potter. • For manual computations, show your complete solution. • For Scilab implementations, include a screenshot of the output. • Submit on or before 22 March 2013, 4:30pm. Late submissions will not be accepted. • If you need clarifications on an item in the problem set, you may email me at nshernandez@dcs.upd.edu.ph or approach me at Rm 319 during consultation hours. ————————————————————————————————————— I. (a) Solve the differential equation given the initial condition: dy −2ty = 2 , dt t +1

y(0) = 4

(b) Express the particular solution you obtained in (a) in the form y = f (t). Write a Scilab script that will plot this f (t) for values of t ranging from t = 0 to t = 2 with a step size of 0.01. (c) Write a Scilab script that uses its ode solver on the given ODE in (a) to obtain a vector of values of the particular solution of the ODE evaluated on the range of values t = 0 to t = 2 with a step size of 0.01 and plot this output. (d) Compare the plot in (b) and (c).

II. The Cauchy-Euler equation is an important example of a linear, secondorder differential equation whose equations are not constant. In particular, it is a differential equation of the form t2 y 00 + pty 0 + qy = 0

(1)

where p and q are real constants and t > 0. (a) Explain why it is reasonable to guess that y = tλ is a solution to equation (1). Show by direct substitution in equation (1) that the guess y − tλ requires λ to be a solution to the characteristic equation λ2 + (p − 1)λ + q = 0

(2)

(b) In the case where equation (2) has two distinct real roots λ1 and λ2 , then the general solution to the Cauchy-Euler equation (1) is y = c1 tλ1 + c2 tλ2 Solve the following Cauchy-Euler initial-value problem: t2 y 00 + 9ty 0 + 12y = 0,

y(1) = 1, y 0 (1) = 0

(c) When equation (2) has a repeated real root λ1 = λ2 = λ, then we have only determined one linearly independent solution (y1 = tλ1 ) of the Cauchy-Euler equation (1). Here we determine a second linearly independent solution. i. Assuming that λ is a repeated root of equation (2), show that 1−p = 2λ. ii. Letting v(t) be an unknown function, consider the guess y2 = vtλ . By direct substitution in the Cauchy-Euler equation, show that v must satisfy the equation tλ t2 v 00 + (2λ + p)tv 0 + (λ2 + (p − 1)λ + q)v = 0 (3) iii. Use your work in (i) and (ii), as well as the fact that λ satisfies the equation λ2 + (p − 1)λ + q = 0 to show that y2 = vtλ is a solution of the Cauchy-Euler equation in the case of a repeated root provided that tv 00 + v 0 = 0

(4)

iv. Show that v(t) = ln t is a solution to equation (4). State the general solution of the Cauchy-Euler equation in the case where the characteristic equation has a single real repeated root.

(d) Solve the following Cauchy-Euler initial-value problem: t2 y 00 − 9ty 0 + 25y = 0,

y(1) = 1, y 0 (1) = 0

(e) When equation (1) has complex roots, say λ1 = a + bi and λ2 = a − bi, then we proceed with a corresponding complex solution to the Cauchy-Euler equation and verify that its real and imaginary parts are themselves real, linearly independent solutions to the equation. In particular, with λ = a+bi, observe that z(t) = tλ = ta+bi = ta tbi . By writing bi )

tbi = eln(t

= ebi ln t

and applying Euler’s formula, show that z(t) = ta [cos(b ln t) + i sin(b ln t)] .

(5)

In addition, show by direct substitution that y1 (t) = ta cos(b ln t) is a solution to the Cauchy-Euler equation when a + bi is a root of the characteristic polynomial. Likewise, show that y2 (t) = ta sin(b ln t) is a solution. State the general solution to the Cauchy-Euler equation in the case where the characteristic polynomial has complex roots λ = a ± bi. (f) Solve the following Cauchy-Euler initial-value problem: t2 y 00 + 3ty 0 + 5y = 0,

y(1) = 1, y 0 (1) = 0

Read more

Similar to

Popular now

Just for you