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1. Ch3: Inverse Laplace Transforms
+
Solution of Linear Differential Equations
Lecture 5

2. Properties of Inverse Laplace Transforms ‫العكسية‬ ‫البالس‬ ‫تحويالت‬ ‫خواص‬
3 ℒ−1
𝑑
𝑑𝑠
𝐹 𝑠 = −𝑡𝑓 𝑡 4 ℒ−1 න
𝑠
∞
𝐹 𝑠 𝑑𝑠 =
𝑓 𝑡
𝑡
5 ℒ−1
𝐹 𝑠 𝑒−𝑎𝑠
= 𝑓 𝑡 − 𝑎 𝑢(𝑡 − 𝑎)
ℒ−1
𝑐1𝐹1(s) ± 𝑐2𝐹2(s) = 𝑐1𝑓1 𝑡 ± 𝑐2𝑓2 𝑡
1 ℒ−1 𝐹 𝑠 ∓ 𝑎 = 𝑒±𝑎𝑡𝑓 𝑡
2
‫المحاضرة‬
‫السابقة‬

3. Find inverse Laplace of
𝐹 𝑠 =
3
𝑠 − 1 2 + 9
𝑒−4𝑠
Solution:
ℒ−1
3
𝑠 − 1 2 + 9
= sin 3𝑡 𝑒𝑡
∴ ℒ−1
3
𝑠 − 1 2 + 9
𝑒−4𝑠 = sin 3 𝑡 − 4 𝑒(𝑡−4)𝑢(𝑡 − 4)
𝐹 𝑠 =
1
𝑠 − 3
𝑒−2𝑠
Solution:
ℒ−1
1
𝑠 − 3
= 𝑒3𝑡
∴ ℒ−1
1
𝑠 − 3
𝑒−2𝑠
= 𝑒3(𝑡−2)
𝑢(𝑡 − 2)

4. Find inverse Laplace of
𝐹 𝑠 =
𝒔
𝒔𝟐 + 2𝒔 + 2
𝑒−3𝑠
Solution:
By applying complete square (‫مربع‬ ‫)إكمال‬
ℒ−1
𝑠
𝑠 + 1 2 + 1
= ℒ−1
𝑠 + 1 − 1
𝑠 + 1 2 + 1
= ℒ−1
𝑠 + 1
𝑠 + 1 2 + 1
− ℒ−1
1
𝑠 + 1 2 + 1
= 𝑒−𝑡 cos 𝑡 − 𝑒−𝑡 sin 𝑡 = 𝑒−𝑡 cos 𝑡 − sin(𝑡)
ℒ−1
𝑠
𝑠 + 1 2 + 1
𝑒−3𝑠
= 𝑒−(𝑡−3)
cos 𝑡 − 3 − sin 𝑡 − 3 𝑢(𝑡 − 3)

5. Find inverse Laplace of
𝑭 𝒔 =
1
𝒔 + 1 𝒔𝟐 + 1
𝑒−𝜋𝑠
Solution:
1
𝒔 + 1 𝒔𝟐 + 1
=
𝑨
𝒔 + 𝟏
+
𝑩𝒔 + 𝑪
𝒔𝟐 + 𝟏
Multiplying both sides by 𝑠 + 1 (𝑠2 + 1) then
Put 𝑠 = −1 → 𝐴 = 0.5, Equating the coefficient of 𝑠2
→ 𝐵 = −0.5
Equating the coefficient of 𝑠0 → 𝐶 = 0.5
ℒ−1
1
𝒔 + 1 𝒔𝟐 + 1
= ℒ−1
0.5
𝑠 + 1
+
−0.5𝑠 + 0.5
𝑠2 + 1
= 0.5𝑒−𝑡 − 0.5 cos 𝑡 + 0.5sin(𝑡)
ℒ−1
1
𝒔 + 1 𝒔𝟐 + 1
𝑒−𝜋𝑠
= 0.5𝑒−(𝑡−𝜋)
− 0.5 cos 𝑡 − 𝜋 + 0.5 sin 𝑡 − 𝜋 𝑢(𝑡 − 𝜋)

6. Solution of linear differential equations
ℒ 𝑦 𝑡 = 𝑌 𝑠
ℒ 𝑦′ 𝑡 = 𝑠𝑌 𝑠 − 𝑦(0)
ℒ 𝑦′′ 𝑡 = 𝑠2𝑌 𝑠 − 𝑠𝑦(0) − 𝑦′(0)
ℒ 𝑦′′′ 𝑡 = 𝑠3𝑌 𝑠 − 𝑠2𝑦 0 − 𝑠𝑦′ 0 − 𝑦′′(0)
and ℒ ‫׬‬
0
𝑡
𝑦 𝑡 𝑑𝑡 =
𝑌(𝑠)
𝒔
Proof:
ℒ 𝑦′ 𝑡 = න
0
∞
𝑦′ 𝑡 𝑒−𝑠𝑡𝑑𝑡 = 𝑦 𝑡 𝑒−𝑠𝑡 ቤ
∞
0
+ 𝑠 න
0
∞
𝑦 𝑡 𝑒−𝑠𝑡𝑑𝑡
∴ ℒ 𝑦′
𝑡 = −𝑦 0 + 𝑠𝑌 𝑠

7. 𝒚′ + 𝒚 = 𝒆−𝒕, 𝒚 𝟎 = 𝟏
Solution:
1
-
‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬
ℒ 𝒚′ + 𝒚 = ℒ 𝑒−𝑡
𝑠𝑌 𝑠 − ถ
𝑦(0)
1
+ 𝑌 𝑠 =
1
𝑠 + 1
𝑠𝑌 𝑠 − 1 + 𝑌 𝑠 =
1
𝑠 + 1
2
-
‫نوجد‬
𝒀(𝒔)
‫في‬
‫طرف‬
𝑠 + 1 𝑌 𝑠 =
1
𝑠 + 1
+ 1 ⇒ 𝑌 𝑠 =
1
𝑠 + 1 2
+
1
𝑠 + 1
3
-
‫ناخذ‬
‫البالس‬ ‫معكوس‬
‫للطرفين‬
ℒ−1 𝑌 𝑠 = ℒ−1
1
𝑠 + 1 2
+
1
𝑠 + 1
‫خواص‬+ ‫الجدول‬
⇒ 𝑦 𝑡 = 𝑡𝑒−𝑡 + 𝑒−𝑡 = (𝑡 + 1)𝑒−𝑡
Solve using Laplace transforms

8. 𝒚′′ + 𝒚 = 𝟏, 𝒚 𝟎 = −𝟏, 𝒚′ 𝟎 = 𝟎
Solution:
1
-
‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬
ℒ 𝒚′′ + 𝒚 = ℒ 1
𝑠2
𝑌 𝑠 − 𝑠 ถ
𝑦(0)
−1
− 𝑦′
(0)
0
+ 𝑌 𝑠 =
1
𝑠
𝑠2𝑌 𝑠 + 𝑠 + 𝑌 𝑠 =
1
𝑠
2
-
‫نوجد‬
𝒀(𝒔)
‫طرف‬ ‫في‬
(𝑠2 + 1)𝑌 𝑠 =
1
𝑠
− 𝑠 =
1 − 𝑠2
𝑠
⇒ 𝑌 𝑠 =
1 − 𝑠2
𝑠(𝑠2 + 1)
3
-
‫ناخذ‬
‫البالس‬ ‫معكوس‬
‫للطرفين‬
ℒ−1 𝑌 𝑠 = ℒ−1
1 − 𝑠2
𝑠(𝑠2 + 1)
𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛
= ℒ−1
ฎ
𝐴
1
𝑠
+
ฎ
𝐵
−2
𝑠 + ฎ
𝐶
0
(𝑠2 + 1)
⇒ 𝑦 𝑡 = 1 − 2𝑐𝑜𝑠 𝑡
Solve using Laplace transforms

9. 𝒚′ − න
0
𝑡
𝑦 𝑡 𝑑𝑡 = 𝒖(𝒕 − 𝟑), 𝒚 𝟎 = 𝟑,
Solution:
1
-
‫كلها‬ ‫للمعادلة‬ ‫البالس‬ ‫ناخذ‬
ℒ 𝒚′ − න
0
𝑡
𝑦 𝑡 𝑑𝑡 = ℒ 𝑢(𝑡 − 3)
𝑠𝑌 𝑠 − ถ
𝑦 0
3
−
𝑌 𝑠
𝑠
=
1
𝑠
𝑒−3𝑠
𝑠 −
1
𝑠
𝑌 𝑠 =
1
𝑠
𝑒−3𝑠 + 3
2
-
‫نوجد‬
𝒀(𝒔)
‫طرف‬ ‫في‬
𝑠2
− 1
𝑠
𝑌 𝑠 =
1
𝑠
𝑒−3𝑠 + 3 ⇒ 𝑌 𝑠 =
1
𝑠2 − 1
𝑒−3𝑠 +
3𝑠
𝑠2 − 1
3
-
‫ناخذ‬
‫البالس‬ ‫معكوس‬
‫للطرفين‬
ℒ−1 𝑌 𝑠 = ℒ−1
1
𝑠2 − 1
𝑒−3𝑠 +
3𝑠
𝑠2 − 1
⇒ 𝑦 𝑡 = sinh 𝑡 − 3 𝑢 𝑡 − 3 + 3cosh 𝑡
Solve using Laplace transforms