Solve the following initial value problem for 1st order PDE SU JUx - 2U₁ = e-U U(x, | U (x, 0) = = 0. 7. Solve the initial value problem Ux+2U₁ =e,t≥0 |U(x, 0) = 1 7. The charateristics are # = = = == they 装 += >* C. are straight lines Along charaterstic Thes. PRE becomes So [u cx, tcx)] = Ux +2 Ut e". du +[e"] еи u Now when t = 0, we have = 〃 = |= = е-и 1 x + fcc) log (x+ feel) (*3) с C = -2x and X = == 50 |= u(x, 0) = log(x+fros) = log (= = +fcol) Exponectinting both Sides, get e = fcc) - = £ + fcc) 들 te plug into (3), get Solution to PDE \\ cxit) = log [x+=+e] = log(x+ + 2x +e) = 109 (±+e) Scanned with CS CamScanner Scanned with CS CamScanner
Solve the following initial value problem for 1st order PDE SU JUx - 2U₁ = e-U U(x, | U (x, 0) = = 0. 7. Solve the initial value problem Ux+2U₁ =e,t≥0 |U(x, 0) = 1 7. The charateristics are # = = = == they 装 += >* C. are straight lines Along charaterstic Thes. PRE becomes So [u cx, tcx)] = Ux +2 Ut e". du +[e"] еи u Now when t = 0, we have = 〃 = |= = е-и 1 x + fcc) log (x+ feel) (*3) с C = -2x and X = == 50 |= u(x, 0) = log(x+fros) = log (= = +fcol) Exponectinting both Sides, get e = fcc) - = £ + fcc) 들 te plug into (3), get Solution to PDE \\ cxit) = log [x+=+e] = log(x+ + 2x +e) = 109 (±+e) Scanned with CS CamScanner Scanned with CS CamScanner
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 5T
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Transcribed Image Text:Solve the following initial value problem for 1st order PDE
SU
JUx - 2U₁ = e-U
U(x,
| U (x, 0) =
= 0.
![7. Solve the initial value problem
Ux+2U₁ =e,t≥0
|U(x, 0) = 1
7.
The charateristics are
# = = =
== they
装
+= >* C.
are straight lines
Along charaterstic Thes. PRE becomes
So
[u cx, tcx)]
=
Ux +2 Ut
e". du
+[e"]
еи
u
Now when t = 0, we have
=
〃
=
|=
=
е-и
1
x + fcc)
log (x+ feel)
(*3)
с
C = -2x
and X = ==
50
|= u(x, 0) = log(x+fros) = log (= = +fcol)
Exponectinting both Sides, get
e
=
fcc)
-
=
£ + fcc)
들 te
plug into (3), get
Solution to PDE
\\ cxit) = log [x+=+e] = log(x+ + 2x +e)
=
109 (±+e)
Scanned with
CS CamScanner
Scanned with
CS CamScanner](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2437b9d-e2ad-49ef-bf1e-5f1a1e22adaa%2F96a59ab6-0cae-423b-84b0-743df5934ba9%2Fzt9y8ci_processed.jpeg&w=3840&q=75)
Transcribed Image Text:7. Solve the initial value problem
Ux+2U₁ =e,t≥0
|U(x, 0) = 1
7.
The charateristics are
# = = =
== they
装
+= >* C.
are straight lines
Along charaterstic Thes. PRE becomes
So
[u cx, tcx)]
=
Ux +2 Ut
e". du
+[e"]
еи
u
Now when t = 0, we have
=
〃
=
|=
=
е-и
1
x + fcc)
log (x+ feel)
(*3)
с
C = -2x
and X = ==
50
|= u(x, 0) = log(x+fros) = log (= = +fcol)
Exponectinting both Sides, get
e
=
fcc)
-
=
£ + fcc)
들 te
plug into (3), get
Solution to PDE
\\ cxit) = log [x+=+e] = log(x+ + 2x +e)
=
109 (±+e)
Scanned with
CS CamScanner
Scanned with
CS CamScanner
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