Exercise 2. (a) Give the definition of solution to the Cauchy problem for a second order linear differential equation with constant coefficients. Consider the second order linear differential equation with constant coefficients y" + ay' + by = g(x), x Є I, with a, bЄ R, I CR is an interval, and g : I → R is a continuous function. The function y = f(x) is a solution to the Cauchy problem if f is twice differentiable and satisfies ƒ ƒ"(x)+af'(x) +bf(x) = g(x) = { ƒ"(x) (x) = Vx Є I, where xo I and yo, y1 E R. (b) Check that (x) = sin² x is a solution over R to y" + 2y = 22 sin² x. Calculating '(x) = 2 sin x cos x, "(x) = 2 cos² x − 2 sin² x and substituting into the equation we obtain - 2 cos² x 2 sin² x2 sin² x = =2 cos² x = - 2(1 − sin² x) from which we conclude that (x) sin² x is a solution on R to the differential equation. (c) Find the general solution over R to the differential equation above. The general solution to the equation on R is the sum of the particular solution (x) above and the solution to the associated homogeneous equation y" + 2y = 0. The latter is solved by means of the characteristic equation x²+2=0 with solutions 1.2 = ±√√√2. The general solution to the homogeneous equation is y = (x) = c₁ sin √√2x + c2 cos √√2x VC1, C2 Є R and the general solution to the inhomogeneous equation is hence y = 4(x) + v(x) = sin² x + c₁ cos √2x + C2 cos √√2x VC1, C2 E R.
Exercise 2. (a) Give the definition of solution to the Cauchy problem for a second order linear differential equation with constant coefficients. Consider the second order linear differential equation with constant coefficients y" + ay' + by = g(x), x Є I, with a, bЄ R, I CR is an interval, and g : I → R is a continuous function. The function y = f(x) is a solution to the Cauchy problem if f is twice differentiable and satisfies ƒ ƒ"(x)+af'(x) +bf(x) = g(x) = { ƒ"(x) (x) = Vx Є I, where xo I and yo, y1 E R. (b) Check that (x) = sin² x is a solution over R to y" + 2y = 22 sin² x. Calculating '(x) = 2 sin x cos x, "(x) = 2 cos² x − 2 sin² x and substituting into the equation we obtain - 2 cos² x 2 sin² x2 sin² x = =2 cos² x = - 2(1 − sin² x) from which we conclude that (x) sin² x is a solution on R to the differential equation. (c) Find the general solution over R to the differential equation above. The general solution to the equation on R is the sum of the particular solution (x) above and the solution to the associated homogeneous equation y" + 2y = 0. The latter is solved by means of the characteristic equation x²+2=0 with solutions 1.2 = ±√√√2. The general solution to the homogeneous equation is y = (x) = c₁ sin √√2x + c2 cos √√2x VC1, C2 Є R and the general solution to the inhomogeneous equation is hence y = 4(x) + v(x) = sin² x + c₁ cos √2x + C2 cos √√2x VC1, C2 E R.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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This is a 3 part question
could you explain part c only please
Transcribed Image Text:Exercise 2.
(a) Give the definition of solution to the Cauchy problem for a second order linear differential equation with constant coefficients.
Consider the second order linear differential equation with constant coefficients
y" + ay' + by = g(x),
x Є I,
with a, bЄ R, I CR is an interval, and g : I → R is a continuous function. The function y = f(x) is a solution to the Cauchy
problem if f is twice differentiable and satisfies
ƒ ƒ"(x)+af'(x) +bf(x) = g(x)
=
{ ƒ"(x) (x) =
Vx Є I,
where xo I and yo, y1 E R.
(b) Check that (x) = sin² x is a solution over R to
y" + 2y = 22 sin² x.
Calculating '(x) = 2 sin x cos x, "(x) = 2 cos² x − 2 sin² x and substituting into the equation we obtain
-
2 cos² x 2 sin² x2 sin² x =
=2 cos² x =
-
2(1 − sin² x)
from which we conclude that (x)
sin² x is a solution on R to the differential equation.
(c) Find the general solution over R to the differential equation above.
The general solution to the equation on R is the sum of the particular solution (x) above and the solution to the associated
homogeneous equation
y" + 2y = 0.
The latter is solved by means of the characteristic equation
x²+2=0
with solutions 1.2 = ±√√√2. The general solution to the homogeneous equation is
y = (x) = c₁ sin √√2x + c2 cos √√2x VC1, C2 Є R
and the general solution to the inhomogeneous equation is hence
y = 4(x) + v(x) = sin² x + c₁ cos √2x + C2 cos √√2x VC1, C2 E R.
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