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Superposition-Differential Equations-Assignemnt and Solution, Exercises of Differential Equations

Dr. Bhim Rao Ambedkar University Differential Equations

This is solved assignment of Differential Equations course. It can be helpful to engineering, computer science, physics and maths students. It was submitted to Prof. Dhanesh Bhatnagar at B R Ambedkar National Institute of Technology. It includes: Superposition, Integer, Antiphase, Input, Response, Periodic, Homogeneous, Roots, Operator

Typology: Exercises

2011/2012

Uploaded on 07/31/2012

dhananjay 🇮🇳

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g(t) = 4 (sin(t)− 1 sin(3t)+ 1 sin(5t)−· · · )

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π 32 52

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A sin(ωt)

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x¨ + ω2 x = A sin(ωt)

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ω2 n

n−ω2

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x¨ + ω2 x = g(t)

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xp = 4 ( sin(t) 1 sin(3t) + )

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ωn

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ω2 32 n−32

nπ n−12 − ω2 · · ·

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ωn

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4 sin(t)

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ωr = 1

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ω

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π ω2

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xp

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n−1

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sin(t)

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ωn

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sin(t)

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ωn

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xp

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2π

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a cos(ωnt) +

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2π

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2π

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ωn

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f(t) = −u(t)t

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f�(t) = −u(t)

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f(t) = u(t)(1 − t)

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f�(t) = −u(t) + δ(t)

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f(t) = (u(t) − u(t − 1))(2t − 1)

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f�(t) = 2(u(t) − u(t − 1)) − δ(t) − δ(t − 1)

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f(t) = (u(t)−u(t−1))t+(u(t−1)−u(t−2))(t−1)+(u(t−1)−u(t−2))(t−2)+ = · · ·

u(t)t − u(t − 1) − u(t − 2) − · · ·

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−1 ± i

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e−t(a cos t + b sin t)

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w(0) = 0

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a = 0

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w(t) = 1

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t > 0

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p(D)x = 1

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xp = 1

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x = 4

1 + e−t(a cos t + b sin t)

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0 = x(0) = 1

4 +a

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a = −4

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x˙ = e−t((−a+b) cos t+(−a−b) sin t)

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0 = x˙(0) = −a + b

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b = −1

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v = 1 u(t)(1 − e−t(cos t + sin t))

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v˙ = −1

4 e−t((−1 + 1) cos t + (−1 − 1) sin t) = 1

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(aD + bI)(2u) = 2aδ(t) + 2bu(t)

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p(D) = 1

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t > 0

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w(t) = t

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a2x¨ + a1x˙ + a0x = 0

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x(0) = 0

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a1 + a0t = 0

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1 = d tt=0 = 1

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p(D) = D2

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w(t) = u(t)t

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a2δ(t) = δ(t)

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Discover Exercises of Differential Equations Dr. Bhim Rao Ambedkar University

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�� g(t) = (^4) π^ (sin(t)− 312 sin(3t)+ 512 sin(5t)−· · · )� �� A sin(ωt) ω^2 ��^ �^ ��^ ��^ x¨ +^ ωn^2 x^ =^ A^ sin(ωt)�^ ��^ ��^ ��^ �^ �� n−ω^2 �� x¨ + ω^2 x = g(t) �� xp = 4 ( (^) ωsin( 2 t)^312 sin(3t)^ + )� �� ωn �� n (^) π (^) n− (^12) n− 32 − (^) ω 2 · · · �� ωn �� ω^4 sin(t) r = 1�^ ��^ ω^ ��^ ��^ ��^1 �^ ��^ ��^ π ω n (^2) − 1 ��^ ��^ xp ��^ �� sin(t) �� ωn �� 1 � �� sin(t)� �� ωn �� xp �� 2 π� �� a cos(ωnt) + b sin(ωnt)� �� (^) ω^2 π n � �� 2 π �� (^) ω^2 π n � �� ωn �� 2 π �� ωn �� f (t) = −u(t)t� f �(t) = −u(t)� �� f (t) = u(t)(1 − t)� f �(t) = −u(t) + δ(t)� �� f (t) = (u(t) − u(t − 1))(2t − 1)� f �(t) = 2(u(t) − u(t − 1)) − δ(t) − δ(t − 1)� �� f (t) = (u(t)−u(t−1))t+(u(t−1)−u(t−2))(t−1)+(u(t−1)−u(t−2))(t−2)+ · · ·= u(t)t − u(t − 1) − u(t − 2) − · · · � �� − 1 ± i� �� e−t(a cos t + b sin t)� �� w(0) = 0 �� w˙ (0+) = 21 � �� a = 0 �� b = 12 � w(t) = 12 u(t)e−t^ sin t� �� t > 0 � �� p(D)x = 1� �� xp = 14 �� x = 41 + e−t(a cos t + b sin t)� �� 0 = x(0) = 14 +a �� a = − 41 � x˙ = e−t((−a+b) cos t+(−a−b) sin t)� �� 0 = x˙ (0) = −a + b �� b = −^14 �� v = 14 u(t)(1 − e−t(cos t + sin t))� �� v˙ = −^14 e−t((− 1 + 1) cos t + (− 1 − 1) sin t) = 12 e−t^ sin t� �� (aD + bI)(2u) = 2aδ(t) + 2bu(t)� �� b = 0 �� a = 21 � p(D) = 12 D� �� t > 0 � w(t) = t �� a 2 x¨ + a 1 x˙ + a 0 x = 0 �� x(0) = 0 �� x˙ (0) = (^) a^12 � �� a 1 + a 0 t = 0 �� a 1 = a 0 = 0� �� 1 = (^) dtd^ t |t (^) =0 = (^) a^12 �� a 2 = 1� �� p(D) = D^2 � �� w(t) = u(t)t� w˙ (t) = u(t) �� w¨(t) = δ(t)� �� a 2 δ(t) = δ(t) �� a 2 = 1�

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�� w(t) �� w¨(t) = 2u(t)� �� w(3)(t) = 2δ(t)� �� a 3 D^3 + a 2 D^2 + a 1 D + a 0 I� t^2 �� a 2 · 2 + a 1 2 t + a 0 t^2 = 0� �� a 0 = a 1 = a 2 = 0� w¨(0) = 2 �� 1

�� a^1 D^3 � u(t)t^2 � 3 =^2 ��^ p(D) =^2 ��^ ��^ ��^ ��^ ��^ w(t) =^ w˙ (t) =^ u(t)2t� w¨(t) = 2u(t)� w(3)(t) = 2δ(t)� �� a 3 w(3)(t) = δ(t) �� a 3 = 21 � �� x(t) = w(t)∗q(t) = ´ (^) t w(t−τ )q(τ ) dτ = ´ (^) t e−k(t−τ^ )^ cos(ωτ ) dτ = e−kt^ ´ (^) t 0 0 0 Re(e(k+iω)τ^ )^ dτ^ = e−ktRee (k+iω)t− 1 k+iω =^ k^2 +^1 ω^2 Re((k^ −^ iω)((cos(ωt)^ −^ e−kt) +^ i^ sin(ωt))^ =^ k^2 +^1 ω^2 (k^ cos(ωt) + ω sin(ωt) − ke−kt)� �� x˙ = (^) k (^2) +^1 ω 2 (−kω sin(ωt) + ω^2 cos(ωt) + k^2 e−kt)� �� x˙ + kx = cos(ωt)� �� x(0) = 0 � �� x(t) = w(t) ∗ q(t) = ´ (^) t w(t − τ )q(τ ) dτ = 1 ´ (^) t 0 ω sin(ωn(t^ −^ τ^ ))^ dτ^ =^ ω^12 cos(ωn(t^ − n (^0) n τ ))|t 0 = (^) ω^12 (1 − cos(ωnt))� �� x˙ = (^) ω^1 sin(ωnt) �� x¨ = cos(ωnt)� �� n n x¨ + ωn^2 x = 1� �� x(0) = 0 �� x˙ (0) = 0 � �� t t (^1 2 1 ) �� t ´ (^2) ∗ t ´

0 (t^ −^ τ ´) (^2) τ dτ ´

0 (t (^2) τ − 2 − tτ 2 + − τ 3 ) dτ = 2 t^43 t^4 + 4 t^4 = 12 t^4 � t t (^1 1 )

t ∗ t^2 = 0 (t − τ )τ 2 dτ = 0 (tτ 2 − τ 3 ) dτ = 3 t^4 4 t^4 = 12 t^4 � �� t ∗ t = t (t − τ )τ dτ = t 0 0 (tτ^ −^ τ^2 )^ dτ^ =^12 t^3 13 t^3 =^16 t^3 � �� (t∗t)∗t = 1 t (t−τ )^3 τ dτ = 1 t 6 0 (t^3 −^3 t^2 τ^ +3tτ^2 −τ^3 )τ^ dτ^ =^1 (^1 3 +^3 1 )t^5 =^1 t^5 � ´ 6 0 6 2

t (^1 1 ) �� t ∗ (t ∗ t) = 16 0 (t − τ )τ 3 dτ = 6 (^14 − 5 )t^5 = 120 t^5 �

Superposition-Differential Equations-Assignemnt and Solution, Exercises of Differential Equations

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