Choose one correct answer from the drop-down multiple choice menus. Consider the following differential equation: The standard form of the given differential equation is (5) + (1) (4) 39 39 dy dx = 20 + 3ey. 3e y=20 (2) +3y= == -20ex (3) -3y=20e² (5) -3y=20e- (6) The integrating factor of the given differential equation is 39 - 3y = -20e 20y=3e - (a) (d) μ(x) = e³ μ(x) = ex (b) μ(x) = e−³ (c) μ(x) = e¯ -20x (e) μ(x) = x (f) μ(x) = ln(x) The solution of the differential equation is given by ÷ A) y(x)=5e+Ce 3 -3x D) y(x)=5e I + Ce³ G) y(x)=-3e+ Ce 5 B) y(x) = 5Ce 3x E) y(x) = -5e -3x + Ce² H) y(x)=5e3Ce C) y(x)=5e-* + Ce² F) y(x)=20e + Ce I) y(x) = 20e 2 +3Ce³z
Choose one correct answer from the drop-down multiple choice menus. Consider the following differential equation: The standard form of the given differential equation is (5) + (1) (4) 39 39 dy dx = 20 + 3ey. 3e y=20 (2) +3y= == -20ex (3) -3y=20e² (5) -3y=20e- (6) The integrating factor of the given differential equation is 39 - 3y = -20e 20y=3e - (a) (d) μ(x) = e³ μ(x) = ex (b) μ(x) = e−³ (c) μ(x) = e¯ -20x (e) μ(x) = x (f) μ(x) = ln(x) The solution of the differential equation is given by ÷ A) y(x)=5e+Ce 3 -3x D) y(x)=5e I + Ce³ G) y(x)=-3e+ Ce 5 B) y(x) = 5Ce 3x E) y(x) = -5e -3x + Ce² H) y(x)=5e3Ce C) y(x)=5e-* + Ce² F) y(x)=20e + Ce I) y(x) = 20e 2 +3Ce³z
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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this is base on diffrenecial equation questions
LINEAR EQUATIONS & EXACT EQUATIONS & SOLUTIONS BY SUBSTITUTIONS
please if you could answer faster i appreaciate of your effort
Transcribed Image Text:Choose one correct answer from the drop-down multiple choice menus.
Consider the following differential equation:
The standard form of the given differential equation is (5) +
(1)
(4)
39
39
2
dy
dx
= 20 + 3ey.
3e y=20
(2)
+3y=
==
-20ex
(3)
-3y= 20e
(5)
-3y=20e-
(6)
The integrating factor of the given differential equation is
39
-
3y = -20e
20y=3e
-
(a)
(d)
μ(x) = e³
μ(x) = ex
(b) μ(x) = e−³
(c) μ(x) = e
-20x
(e) μ(x) = x
(f) μ(x) = ln(x)
The solution of the differential equation is given by
÷
A)
y(x)=5e+Ce 3
-3x
D) y(x)=5e
I
+ Ce³
G)
y(x)=-3e+ Ce 5
B)
y(x) = 5Ce
3x
E)
y(x) = -5e
-3x
+ Ce²
H)
y(x)=5e3Ce
C) y(x)=5e-* + Ce²
F) y(x)=20e + Ce
I)
y(x) = 20e
2
+3Ce³
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