Solve the folowing differential equations using Laplace transform method.(1) y"+4y'+5y = 50, y(0) = 5, y'(0) =-5(2) y"+16y =48(t –7), y(0) =-1, y'(0) =0(3) y"-y'-2y = 12u (1 -7)sint, y (0) = 1, y'(0) =-1%3D(4) y"+3y'+2y =2u (t –2), y(0)=0, y'(0) =0(5) y"+3y'+2y =f (t). y(0)=1, y'(0) = 0where f (t)=0, 0t(6) y"+5y'+6y =u (1 – x)cos(1 - 7), y (0) =1, y'(0) =0(7) y"-5y'+6y =tu(t -1), y (0)=0, y'(0) =0(8) y"-5y'+6y =l+tu(t -2). y(0)=0, y '(0)=1(9) y"-5y'+10y =tu(1 -3), y(0) =1, y'(0) = 1(10) y"+2y'+10y =e*u (t -1), y (0) =-1, y'(0) = 0(11) y"-y'-2y =e u(1-1). y(0) =1, y'(0) = 0(12) y"-3y'+2y = 4e, y (0) =-3, y'(0) = 5(13) y"-y'-2y 20sin 2r, y(0)=1, y'(0)=2H y"+3y'+2y =2(1² +1 +1), y(0) =2, y'(0) = 0(15) y"+2y'+5y =e*sint, y(0) = 0, y'(0) =1(16) y +Jydt =1-e(17)+2y +Jydt sint, y(0)=1%3Dd'y(18)dt+2y =18(1-1). y (0) =0, y'(0) =0dt0<1 <1(19) y"+4y =f (t), y(0)=0, y'(0) =1 where f(t)=t>1

(a) Given, y''+3y'+2y=2(t2+t+1) and the conditions are y(0)=2, y'(0)=0 Now, y''+3y'+2y=2(t2+t+1)…

Solve the folowing differential equations using Laplace transform method. (1) y"+4y'+5y =50, y(0)=5, y '(0) =-5 (2) y"+16y = 48(t -z), y(0) =-1, y '(0) = 0 (3) y"-y'-2y = 12u (1 – z)sint, y (0) = 1, y'(0) =-1 (4) y"+3y'+2y =2u (t –2), y(0)=0, y'(0) =0 (5) y"+3y'+2y =f (t), y(0)=1, y (0) = 0 %3D %3D where f(1)=0, 0 (6) y"+5y'+6y =u (1 – z)cos(t – 7), y (0) =1, y'(0) = 0 (7) y"-5y'+6y =tu (t-1), y (0)=0, y'(0)=0 (8) y"-5y'+6y =1+fu(t -2). y(0)=0, y'(0) =1 (9) y"-5y'+10y =tu(t – 3). y(0) =1, y'(0) =1 (10) y"+2y'+10y =e*u (t –1), y (0) = -1, y'(0) =0 (11) y"-y'-2y =e u (t -1), y (0)=1, y'(0) =0 (12) y"-3y'+2y = 4e, y(0)=-3, y'(0) = 5 (13) y"-y'-2y =20sin 2r, y(0)=1, y'(0)=2 y"+3y'+2y =2(1 +1 +1). y(0) =2, y'(0) =0 %3D (15) y"+2y'+5y =e*sint, y(0)=0, y'(0) =1 (16) y +ydt = 1-e (17) dt dy +2y +[ydt = sint, y(0)=1 d'ydy di +2y =t8(1-1). y (0) =0, y'(0)=0 (18 +3 dt 0<1 <1 (19) y"+4y =f (). y(0)=0, y'(0) =1 where f()=- t>1

Solve the folowing differential equations using Laplace transform method. (1) y"+4y'+5y =50, y(0)=5, y '(0) =-5 (2) y"+16y = 48(t -z), y(0) =-1, y '(0) = 0 (3) y"-y'-2y = 12u (1 – z)sint, y (0) = 1, y'(0) =-1 (4) y"+3y'+2y =2u (t –2), y(0)=0, y'(0) =0 (5) y"+3y'+2y =f (t), y(0)=1, y (0) = 0 %3D %3D where f(1)=0, 0 (6) y"+5y'+6y =u (1 – z)cos(t – 7), y (0) =1, y'(0) = 0 (7) y"-5y'+6y =tu (t-1), y (0)=0, y'(0)=0 (8) y"-5y'+6y =1+fu(t -2). y(0)=0, y'(0) =1 (9) y"-5y'+10y =tu(t – 3). y(0) =1, y'(0) =1 (10) y"+2y'+10y =eu (t –1), y (0) = -1, y'(0) =0 (11) y"-y'-2y =e u (t -1), y (0)=1, y'(0) =0 (12) y"-3y'+2y = 4e, y(0)=-3, y'(0) = 5 (13) y"-y'-2y =20sin 2r, y(0)=1, y'(0)=2 y"+3y'+2y =2(1 +1 +1). y(0) =2, y'(0) =0 %3D (15) y"+2y'+5y =esint, y(0)=0, y'(0) =1 (16) y +ydt = 1-e (17) dt dy +2y +[ydt = sint, y(0)=1 d'ydy di +2y =t8(1-1). y (0) =0, y'(0)=0 (18 +3 dt 0<1 <1 (19) y"+4y =f (). y(0)=0, y'(0) =1 where f()=- t>1

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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Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

Solve the folowing differential equations using Laplace transform method.
(1) y"+4y'+5y = 50, y(0) = 5, y'(0) =-5
(2) y"+16y =48(t –7), y(0) =-1, y'(0) =0
(3) y"-y'-2y = 12u (1 -7)sint, y (0) = 1, y'(0) =-1
%3D
(4) y"+3y'+2y =2u (t –2), y(0)=0, y'(0) =0
(5) y"+3y'+2y =f (t). y(0)=1, y'(0) = 0
where f (t)=0, 0<t <n; f(t)=sin 21, t>t
(6) y"+5y'+6y =u (1 – x)cos(1 - 7), y (0) =1, y'(0) =0
(7) y"-5y'+6y =tu(t -1), y (0)=0, y'(0) =0
(8) y"-5y'+6y =l+tu(t -2). y(0)=0, y '(0)=1
(9) y"-5y'+10y =tu(1 -3), y(0) =1, y'(0) = 1
(10) y"+2y'+10y =e*u (t -1), y (0) =-1, y'(0) = 0
(11) y"-y'-2y =e u(1-1). y(0) =1, y'(0) = 0
(12) y"-3y'+2y = 4e, y (0) =-3, y'(0) = 5
(13) y"-y'-2y 20sin 2r, y(0)=1, y'(0)=2
H y"+3y'+2y =2(1² +1 +1), y(0) =2, y'(0) = 0
(15) y"+2y'+5y =e*sint, y(0) = 0, y'(0) =1
(16) y +Jydt =1-e
(17)
+2y +Jydt sint, y(0)=1
%3D
d'y
(18)
dt
+2y =18(1-1). y (0) =0, y'(0) =0
dt
0<1 <1
(19) y"+4y =f (t), y(0)=0, y'(0) =1 where f(t)=
t>1

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