Six second-order equations and four y(t)-graphs are given below. For each y(t)-graph, determine the second-order equation for which y(t) is a solution, and state briefly how you know your choice is correct. You should do this exercise WITHOUT using technology. d²y d (i) +16y=10 dt² +16y=-10 + 16y = 5 cos(3t) dt² d'y d²y (iv) +14y = 2 cos(4t) +16y= cos(4t) = +2dy + 16y= dt2 dt (a) (b) (v) -2+ (vi) AM cos(4t)

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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I attached the soloution that u gave for first 3 parts they said to find the solution of y(t) And also state briefly how your solouution is correct uh need go find all parts and plz have a look what they said to solve plz Also don't reject this plz
Six second-order equations and four y(t)-graphs are given below. For each y(t)-graph, determine the second-order
equation for which y(t) is a solution, and state briefly how you know your choice is correct. You should
do this exercise WITHOUT using technology.
d²y
d
(i) +16y=10
dt²
+16y=-10
+ 16y = 5 cos(3t)
d'y
d²y
(iv)
+ 14y = 2 cos(4t)
+ 16y= = cos(4t)
+2dy +16y=
dt2
dt
(b)
-2+
C
m
-2+
म
2+
(d)
Wi
27
y(t)-graphs for Exercise 3.
(vi)
AMA
cos(4t)
Transcribed Image Text:Six second-order equations and four y(t)-graphs are given below. For each y(t)-graph, determine the second-order equation for which y(t) is a solution, and state briefly how you know your choice is correct. You should do this exercise WITHOUT using technology. d²y d (i) +16y=10 dt² +16y=-10 + 16y = 5 cos(3t) d'y d²y (iv) + 14y = 2 cos(4t) + 16y= = cos(4t) +2dy +16y= dt2 dt (b) -2+ C m -2+ म 2+ (d) Wi 27 y(t)-graphs for Exercise 3. (vi) AMA cos(4t)
Answer)
i) day +168 = 10
df 2
The associated homogeneous
equation is:
d²y +16y=0
The auxiliary equation of the homogeneous differential
equation
m² +16=0
m = ± 24
Therefore, the complementary solution of the equation is:
y₁ = Acosq++ Bsin4t
&p=a
particular solution as:
Assume the
with respect tot: y = 0
wo times differenziati
substitute Jp any yo in cquation then we will get,
ур
0 +16a=10
Then yp=&
the general solution of the equation @is: y = GH+ Yp
[4 = Acos 4t + Bgn 4+ + 5/8.
i) day
de ² +169 = -10
The particuler solution for
yp = - F
Therefore, the
solution of the given equation is:
y = A los 4t + 1asin 4t - 5/8
+164 = 5 cas (36)
the particular solution for given differential equal
Vp = _gacos (36)-9b3in (34)
y'p in given equation:
and
cos (st) - 9bsin(31) + 16 (acos (at) + bsin (3+) = 5 cas (36)
(-99 +16 a ) cos( 3(-) + (-96-+166) Sin (3+) = 5 cos (34)
7a (os (31) + 7bsin (at) = 5 (os(3+)
a=5/7, b=0
we get,
Therefore, the
Particular solution is yp = 5/7 (as (34)
The solution of the given D.E is
m) day
Assume
2 على
given
the x differential equation is 8
is yp = acos (36) + bsin (at)
substituti yp
y = A los 4+ + B sin4+ + 5/7 cos (36).
Transcribed Image Text:Answer) i) day +168 = 10 df 2 The associated homogeneous equation is: d²y +16y=0 The auxiliary equation of the homogeneous differential equation m² +16=0 m = ± 24 Therefore, the complementary solution of the equation is: y₁ = Acosq++ Bsin4t &p=a particular solution as: Assume the with respect tot: y = 0 wo times differenziati substitute Jp any yo in cquation then we will get, ур 0 +16a=10 Then yp=& the general solution of the equation @is: y = GH+ Yp [4 = Acos 4t + Bgn 4+ + 5/8. i) day de ² +169 = -10 The particuler solution for yp = - F Therefore, the solution of the given equation is: y = A los 4t + 1asin 4t - 5/8 +164 = 5 cas (36) the particular solution for given differential equal Vp = _gacos (36)-9b3in (34) y'p in given equation: and cos (st) - 9bsin(31) + 16 (acos (at) + bsin (3+) = 5 cas (36) (-99 +16 a ) cos( 3(-) + (-96-+166) Sin (3+) = 5 cos (34) 7a (os (31) + 7bsin (at) = 5 (os(3+) a=5/7, b=0 we get, Therefore, the Particular solution is yp = 5/7 (as (34) The solution of the given D.E is m) day Assume 2 على given the x differential equation is 8 is yp = acos (36) + bsin (at) substituti yp y = A los 4+ + B sin4+ + 5/7 cos (36).
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