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Laplace Transformation and Differential equations Mongi BLEL King Saud University March 17, 2020 Mongi BLEL Laplace Transformation and Differential equations Exercise Use the Laplace transform to solve the initial-value problem y + y = e

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Laplace Transformation and Differential equations

Mongi BLEL King Saud University

March 17, 2020

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Use the Laplace transform to solve the initial-value problem y′ + y = e−x + ex + cos x + sin x, y(0) = 1.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 1 : We begin by taking the Laplace transform of both sides to achieve L

  • y′

+ L[y] = L(e−x + ex + cos x + sin x). We know that L(e−x + ex + cos x + sin x) = 1 s + 1 + 1 s − 1 + s s2 + 1 + 1 s2 + 1. Denote Y (s) = L[y], then Y (s) = 1 s + 1 + 1 (s + 1)2 + 1 2( 1 s − 1 − 1 s + 1) + s (s + 1)(s2 + 1) + 1 (s + 1)(s2 + 1) = 1 2(s + 1) + 1 (s + 1)2 + 1 2(s − 1) + 1 s2 + 1. The solution of the differential equation: y = 1 2e−x + xe−x + 1 2ex + sin x.

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Use Laplace transforms to solve the initial-value problem y′ − 2y = 5 + cos x + e2x + e−x, y(0) = 4.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 2 : By taking the Laplace transform of both sides of the differential equation, we get: sY − 4 − 2Y = 5 s + s s2 + 1 + 1 s − 2 + 1 s + 1. Then Y (s) = 4 s − 2 + 5 s(s − 2) + s (s − 2)(s2 + 1) + 1 (s − 2)2 + 1 (s + 1)(s − 2) = 4 s − 2 + 5 2

  • 1

s − 2 − 1 s

  • + 2

5 1 s − 2 + 1 5 −2s + 1 s2 + 1 + 1 (s − 2)2 + 1 3 1 s − 2 − 1 3 1 s + 1 = 217 30(s − 2) − 5 2s − 2 5 s s2 + 1 + 1 5 1 s2 + 1 + 1 (s − 2)2 − 1 3 1 s + 1

Mongi BLEL

Laplace Transformation and Differential equations

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and y(x) = 217 30 e2x − 5 2 − 2 5 cos x + 1 5 sin x + xe2x − 1 3e−x.

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Use the Laplace transform to solve the initial-value problem y′′ − 2y′ + 2y = cos x, y(0) = 1, y′(0) = −1

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 3 : Using the Laplace transform of both sides of the differential equation, we get: s2Y (s) − s + 1 − 2(sY (s) − 1) + 2Y (s) = s s2 + 1. Then Y (s) = s − 3 (s − 1)2 + 1 + s (s2 + 1)((s − 1)2 + 1). Solving for Y (s), we find: Y (s) = s − 1 (s − 1)2 + 1 − 2 (s − 1)2 + 1 + 1 5 s s2 + 1 − 2 5 1 s2 + 1 −1 5 s − 1 (s − 1)2 + 1 + 3 5 1 (s − 1)2 + 1. Then y = ex cos x − 2ex sin x + 1 5 cos x − 2 5 sin x − 1 5ex cos x + 3 5ex sin x = 4 5ex cos x − 7 5ex sin x + 1 5 cos x − 2 5 sin x

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Use Laplace transforms to solve the initial-value problem y′ + y = 5H(x − 1) + exH(x − 1) + H(x − 1) cos x, y(0) = 2.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 4 : Using the Laplace transform of both sides of the differential equation, we get: (s + 1)Y − 2 = L [5H(x − 1) + exH(x − 1) + H(x − 1) cos x] . Since L[H(x − 1)] = 1 s e−s, L[exH(x − 1)] = e−(s−1) s − 1 and L[H(x − 1) cos x] = e−s sin 1 + cos 1 s2 + 1 . We have: Y (s) = 2 s + 1 + 5e−s s(s + 1) + e−(s−1) (s − 1)(s + 1) + (sin 1 + cos 1)e−s (s + 1)(s2 + 1) = 2 s + 1 + 5e−s s − e−s s + 1 + 1 2 e−(s−1) s − 1 −1 2 e−(s−1) s + 1 + (sin 1 + cos 1) 2 e−s s + 1 −(sin 1 + cos 1) 2 se−s s2 + 1 + (sin 1 + cos 1) 2 e−s s2 + 1

Mongi BLEL

Laplace Transformation and Differential equations

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We have L−1

4 s+1

  • = 4e−x,

L−1 5 s e−s

  • = 5H(x − 1) and

L−1

  • 5

s + 1e−s

  • = 5H(x − 1)e−(x−1). Then

y(x) = 4e−x + 5H(x − 1) − 5H(x − 1)e−(x−1).

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Use the Laplace transform to solve the initial-value problem y′′ + 2y′ + 5y = H(x − 2), y(0) = 1, y′(0) = 0

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 5 : Using the Laplace transform of both sides of the differential equation, we get: s2Y (s) − sy(0) − y′(0) + 2(sY (s) − y(0)) + 5Y (s) = e−2s s . Then Y (s)

  • s2 + 2s + 5
  • = s + 2 + e−2s

s . Solving for Y (s), we find: Y (s) = s + 2 s2 + 2s + 5 + e−2s 1 s (s2 + 2s + 5). We have s + 2 s2 + 2s + 5 = s + 1 (s + 1)2 + 4 + 1 (s + 1)2 + 4 and L−1

  • s + 1

(s + 1)2 + 4

  • = e−x cos(2x) and

L−1

  • 2

(s + 1)2 + 4

  • = e−x sin(2x).

Mongi BLEL

Laplace Transformation and Differential equations

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e−2s s (s2 + 2s + 5) = 1 5e−2s 1 s − (s + 1) + 1 (s + 1)2 + 4

  • =

1 5e−2s 1 s − s + 1 (s + 1)2 + 4 + 1 (s + 1)2 + 4

  • .

L−1

  • e−2s

1 s − s + 2 s2 + 2s + 5

  • =

H(x − 2)[1 −e−(x−2)

  • cos 2(x − 2) + 1

2 sin 2(x − 2)

  • ].

The solution y(x) to the initial-value problem is y(x) = e−x cos(2x) + e−x 2 sin(2x) + 1 5H(x − 2)[1 −e−(x−2) cos 2(x − 2) + e−(x−2) 2 sin 2(x − 2)]

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Solve the differential equation y′ + 3y = 13 sin(2t), y(0) = 6.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 6 : We take the Laplace transform of each member of the differential equation: L(y′)+3L(y) = 13L(sin(2t)). Then (s +3)Y (s)−6 = 6+ 26 s2 + 4. Y (s) = 6 s + 3 + 26 (s + 3)(s2 + 4) = 6 s + 3 + −2s + 6 s2 + 4 and y = 6L−1( 1 s + 3) − 2L−1( s s2 + 4) + 6L−1( 1 s2 + 4) = 6e−3t − 2 cos(2t) + 3 sin(2t).

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Solve the differential equation y

′′ − 3y′ + 2y = e−4t,

y(0) = 1, y′(0) = 5.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 7 : We take the Laplace transform of each member of the differential equation: L(y

′′) − 3L(y′) + 2L(y) = L(e−4t). Then

Y (s) = s2 + 6s + 9 (s − 1)(s − 2)(s + 4 and y = −16 5 et + 25 5 e2t + 1 30e−4t.

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Solve the differential equation y

′′ − 6y′ + 9y = t2e3t,

y(0) = 2, y′(0) = 17. Solution 8 : We take the Laplace transform of each member of the differential equation: Y (s) = 2s + 5 (s − 3)2 + 2 (s − 3)5 . y = 2e3t + 11te3t 1

12t4e3t.

Mongi BLEL

Laplace Transformation and Differential equations

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Exercise Solve the following differential equation: y′ − 2y = f (x), withy(0) = 3, f (x) = 3 cos x for x ≥ 1 and f (x) = 0, for 0 ≤ x ≤ 1.

Mongi BLEL

Laplace Transformation and Differential equations

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Solution 9 : L (f (x)) = − 3s

s2+1e−s. Then sF(s) − 3 − 2F(s) = − 3s s2+1e−s and

F(s) = 1 s − 2

  • 3 −

3s s2 + 1e−s

  • =

3 s − 2 + 6 5 s s2 + 1e−s − 3 5 1 s2 + 1e−s. L−1

3 s−2

  • = 3e2t,

L−1

6 5 s s2+1e−s

= 6

5 cos(t − 1)H(t − 1),

L−1

3 5 1 s2+1e−s

= 3

5 sin(t − 1)H(t − 1).

y(t) = 3e2t + 6 5 cos(t − 1)H(t − 1) − 3 5 sin(t − 1)H(t − 1).

Mongi BLEL

Laplace Transformation and Differential equations

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Solve the initial value problem y′′ + y′ + y = sin(x), y(0) = 1, y′(0) = −1.

Mongi BLEL

Laplace Transformation and Differential equations

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L{y′(x)} = sY (s)−y(0) = sY (s)−1, L{y′′(x)} = s2Y (s)−sy(0)−y′(0) Taking Laplace transforms of the differential equation, we get (s2 + s + 1)Y (s) − s = 1 s2 + 1. Then Y (s) = s s2 + s + 1 + 1 (s2 + s + 1)(s2 + 1). s s2 + s + 1 = s (s + 1

2)2 + 3 4

= s + 1

2

(s + 1

2)2 + ( 1 2

√ 3)2 − 1 √ 3

1 2

√ 3 (s + 1

2)2 + ( 1 2

Mongi BLEL

Laplace Transformation and Differential equations

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Finding the inverse Laplace transform. Since s s2 + s + 1 = s (s + 1

2)2 + 3 4

= s + 1

2

(s + 1

2)2 + ( 1 2

√ 3)2 − 1 √ 3

1 2

√ 3 (s + 1

2)2 + ( 1 2

√ 3) we have L−1

  • s

s2 + s + 1

  • = e− 1

2 x cos(1

2 √ 3x) − 1 √ 3 e− 1

2 x sin(1

2 √ 3x). Also, we have 1 (s2 + s + 1)(s2 + 1) = s + 1 s2 + s + 1 − s s2 + 1, 1 s2 + s + 1 = 1 (s + 1

2)2 + 3 4

, and L−1

  • 1

s2 + s + 1

  • =

2 √ 3 e− 1

2 x sin(1

2 √ 3x).

Mongi BLEL

Laplace Transformation and Differential equations

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Then L−1

  • 1

(s2 + s + 1)(s2 + 1)

  • = L−1
  • s

s2 + s + 1

  • +L−1
  • 1

s2 + s + 1

  • −L

We obtain y(x) = 2e− 1

2 x cos(1

2 √ 3x) − cos(x).

Mongi BLEL

Laplace Transformation and Differential equations

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Solve the system of linear differential equation: dx

dt

= −2x + y,

dy dt

= x − 2y with the initial conditions x(0) = 1, y(0) = 2.

Mongi BLEL

Laplace Transformation and Differential equations

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Taking the Laplace transform of the equations, we get sX(s) − 1 = −2X(s) + Y (s), sY (s) − 2 = X(s) − 2Y (s), where X(s) = L{x(x)}, Y (s) = L{y(x)}. then (s + 2)X(s) − Y (s) = 1, −X(s) + (s + 2)Y (s) = 2 The solutions of the linear system of equations on X and Y are X(s) =

s+4 s2+4s+3,

Y (s) =

2s+5 s2+4s+3.

Mongi BLEL

Laplace Transformation and Differential equations

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Using the inverse Laplace transform, we have s + 4 (s + 1)(s + 3) =

3 2

s + 1 −

1 2

s + 3, 2s + 5 (s + 1)(s + 3) =

3 2

s + 1 +

1 2

s + 3, we obtain x(x) = 3 2e−t − 1 2e−3t, y(x) = 3 2e−t + 1 2e−3t.

Mongi BLEL

Laplace Transformation and Differential equations