Laplace Transformation and Differential equations Mongi BLEL King Saud University March 17, 2020 Mongi BLEL Laplace Transformation and Differential equations
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