Mongi BLEL King Saud University - KSU

Mongi BLEL
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
Solution 1 : We begin by taking the Laplace transform of both sides to achieve
L [ y ′ ]
We know that
s + 1 +
Y (s) = 1
y = 1
1
Exercise
y ′ − 2y = 5 + cos x + e2x + e−x , y(0) = 4.
Mongi BLEL Laplace Transformation and Differential equations
Solution 2 : By taking the Laplace transform of both sides of the differential
equation, we get: sY − 4− 2Y = 5
s +
s
and
3 e−x .
Exercise
y ′′ − 2y ′ + 2y = cos x , y(0) = 1, y ′(0) = −1
Mongi BLEL Laplace Transformation and Differential equations
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 .
(s − 1)2 + 1 +
Y (s), we find:
Y (s) = s − 1
(s − 1)2 + 1 − 2
5 cos x − 2
5 sin x − 1
5 ex cos x +
5 ex sin x +
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
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:
(s + 1)(s2 + 1)
We have L−1 [
y(x) = 4e−x + 5H(x − 1)− 5H(x − 1)e−(x−1).
Mongi BLEL Laplace Transformation and Differential equations
Exercise
y ′′ + 2y ′ + 5y = H(x − 2), y(0) = 1, y ′(0) = 0
Mongi BLEL Laplace Transformation and Differential equations
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 .
) = s + 2 +
e−2s
find:
1
e−2s
y(x) = e−x cos(2x) + e−x
2 sin(2x) +
2 sin 2(x − 2)]
Exercise
Mongi BLEL Laplace Transformation and Differential equations
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 .
Mongi BLEL Laplace Transformation and Differential equations
Exercise
y ′′ − 3y ′ + 2y = e−4t , y(0) = 1, y ′(0) = 5.
Mongi BLEL Laplace Transformation and Differential equations
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 +
Exercise
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
4e3t .
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
Solution 9 : L (f (x)) = − 3s
s2+1 e−s . Then sF (s)− 3− 2F (s) = − 3s
s2+1 e−s and
F (s) = 1
5
1
L−1 ( 3 5
y(t) = 3e2t + 6
5 sin(t − 1)H(t − 1).
Solve the initial value problem
y ′′ + y ′ + y = sin(x), y(0) = 1, y ′(0) = −1.
Mongi BLEL Laplace Transformation and Differential equations
L{y ′(x)} = sY (s)−y(0) = sY (s)−1, L{y ′′(x)} = s2Y (s)−sy(0)−y ′(0) = s2Y (s)−s+1.
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 +
Finding the inverse Laplace transform. Since
s
Then
1
2
Mongi BLEL Laplace Transformation and Differential equations
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
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 .
s + 4