h(x) = cF' (x) – cGʻ(x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a)+cG(a), and from here solve a linear system to show that
h(x) = cF' (x) – cGʻ(x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a)+cG(a), and from here solve a linear system to show that
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
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Solve 3 part only and take a thumb up plz solve this perfectly.
![Question 1. Consider the second order partial differential equation
(1)
for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is
a constant.
1. For any twice differentiable functions F =
satisfies 1).
F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct)
Partial differential equations such as 1)
the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are
given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h.
are often solved as initial value problems, where the initial description of
2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If
u = u(t, x) solves the initial value problem described above, show that
g(x) = F(x) + G(x)
and
h(x) = cF' (x) – cG' (x).
3. By integrating the last equation for h(x), show that for any constant a E R,
| h(s) ds = cF(x) – cG(x) – cF(a) + cG(a),
and from here solve a linear system to show that
F(2) = (o(2) +[ h(0) ds + F(a) – G(a).
F(x) =
and
G(x) = ; (9(=) – h(s) ds – F(a) + G(a)
4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial
value problem for 1):
1
1
(r+ct
u(t, æ) = ; [g(x + et) + g(x – ct)] + h(s)ds.
2c
r-ct
xye-(2²+v°+z*). Find all points (x, y, z) such that Vf(x, y, z) =
Question 2. Consider the function f(x, y, z)
(0,0,0).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b6ea445-fcb8-43e5-8ecc-e13cb9c0bb44%2F73cabec6-97f2-4a9b-a7f4-4726d71725e1%2Fdvribzp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 1. Consider the second order partial differential equation
(1)
for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is
a constant.
1. For any twice differentiable functions F =
satisfies 1).
F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct)
Partial differential equations such as 1)
the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are
given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h.
are often solved as initial value problems, where the initial description of
2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If
u = u(t, x) solves the initial value problem described above, show that
g(x) = F(x) + G(x)
and
h(x) = cF' (x) – cG' (x).
3. By integrating the last equation for h(x), show that for any constant a E R,
| h(s) ds = cF(x) – cG(x) – cF(a) + cG(a),
and from here solve a linear system to show that
F(2) = (o(2) +[ h(0) ds + F(a) – G(a).
F(x) =
and
G(x) = ; (9(=) – h(s) ds – F(a) + G(a)
4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial
value problem for 1):
1
1
(r+ct
u(t, æ) = ; [g(x + et) + g(x – ct)] + h(s)ds.
2c
r-ct
xye-(2²+v°+z*). Find all points (x, y, z) such that Vf(x, y, z) =
Question 2. Consider the function f(x, y, z)
(0,0,0).
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