Solve the equation 3uy + Uxy = 0. (Hint: Let v = uy.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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[Partial
The second picture is for context.
![2. Solve the equation 3uy + xy
=
= 0. (Hint: Let v =
=Uy.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45b359f0-fb4a-47fb-b3b7-86deec526e99%2F7e8cad89-1c73-4080-8958-64e10d130af9%2Fro2nnn_processed.png&w=3840&q=75)
Transcribed Image Text:2. Solve the equation 3uy + xy
=
= 0. (Hint: Let v =
=Uy.)
![THE VARIABLE COEFFICIENT EQUATION
The equation
ux + yuy=0
(4)
is linear and homogeneous but has a variable coefficient (y). We shall illustrate
for equation (4) how to use the geometric method somewhat like Example 1.
The PDE (4) itself asserts that the directional derivative in the direction
of the vector (1, y) is zero. The curves in the xy plane with (1, y) as tangent
vectors have slopes y (see Figure 3). Their equations are
This ODE has the solutions
dy y
dx
y = Ce*.
(6)
These curves are called the characteristic curves of the PDE (4). As C is
changed, the curves fill out the xy plane perfectly without intersecting. On
each of the curves u(x, y) is a constant because
d
ди
ди
-u(x, Ce*) = + Ce = Ux + yuy
ay
dx
əx
It follows that
Thus u(x, Cet) = u(0, Ceº) = u(0, C) is independent of x. Putting y = Ce
and C = e-*y, we have
u(x, y) = u(0, e *y).
(5)
= 0.
Figure 3
u(x, y) = f(e *y)
(7)
is the general solution of this PDE, where again f is an arbitrary function
of only a single variable. This is easily checked by differentiation using
the chain rule (see Exercise 4). Geometrically, the "picture" of the solution
u(x, y) is that it is constant on each characteristic curve in Figure 3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45b359f0-fb4a-47fb-b3b7-86deec526e99%2F7e8cad89-1c73-4080-8958-64e10d130af9%2Fvymgfyi_processed.png&w=3840&q=75)
Transcribed Image Text:THE VARIABLE COEFFICIENT EQUATION
The equation
ux + yuy=0
(4)
is linear and homogeneous but has a variable coefficient (y). We shall illustrate
for equation (4) how to use the geometric method somewhat like Example 1.
The PDE (4) itself asserts that the directional derivative in the direction
of the vector (1, y) is zero. The curves in the xy plane with (1, y) as tangent
vectors have slopes y (see Figure 3). Their equations are
This ODE has the solutions
dy y
dx
y = Ce*.
(6)
These curves are called the characteristic curves of the PDE (4). As C is
changed, the curves fill out the xy plane perfectly without intersecting. On
each of the curves u(x, y) is a constant because
d
ди
ди
-u(x, Ce*) = + Ce = Ux + yuy
ay
dx
əx
It follows that
Thus u(x, Cet) = u(0, Ceº) = u(0, C) is independent of x. Putting y = Ce
and C = e-*y, we have
u(x, y) = u(0, e *y).
(5)
= 0.
Figure 3
u(x, y) = f(e *y)
(7)
is the general solution of this PDE, where again f is an arbitrary function
of only a single variable. This is easily checked by differentiation using
the chain rule (see Exercise 4). Geometrically, the "picture" of the solution
u(x, y) is that it is constant on each characteristic curve in Figure 3.
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