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Problem Set 5

The document presents solutions to three differential equations using Laplace transforms, detailing the steps taken to derive the solutions. The first equation is solved to yield y(t) = -1/2 e^(-t) + 1/2 e^(t) = sinh(t). The second and third equations involve more complex manipulations, ultimately leading to expressions for Y(s) and their inverse transforms.

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İlay Öztürk
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13 views5 pages

Problem Set 5

The document presents solutions to three differential equations using Laplace transforms, detailing the steps taken to derive the solutions. The first equation is solved to yield y(t) = -1/2 e^(-t) + 1/2 e^(t) = sinh(t). The second and third equations involve more complex manipulations, ultimately leading to expressions for Y(s) and their inverse transforms.

Uploaded by

İlay Öztürk
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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EEF 210E DIFFERENTIAL EQUATIONS

PROBLEM SET 5

1. Solve the differential equation 𝒚′′′ − 𝟐𝒚′′ − 𝒚′ + 𝟐𝒚 = 𝟎 via Laplace transform where 𝒚(𝟎) = 𝟎, 𝒚′ (𝟎) = 𝟏,
and 𝒚′′ (𝟎) = 𝟎.

𝕃{𝑦 ′′′ − 2𝑦 ′′ − 𝑦 ′ + 2𝑦}(𝑠) = 0

(𝑠 3 𝑌(𝑠) − 𝑠 2 𝑦(0) − 𝑠𝑦 ′ (0) − 𝑦′′(0)) − 2(𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0)) − (𝑠𝑌(𝑠) − 𝑦(0)) + (2𝑌(𝑠)) = 0

⇒ (𝑠 3 − 2𝑠 2 − 𝑠 + 2)𝑌(𝑠) − (𝑠 2 − 2𝑠 − 1)𝑦(0) − (𝑠 − 2)𝑦 ′ (0) − (1)𝑦 ′′ (0) = 0


⇒ (𝑠 3 − 2𝑠 2 − 𝑠 + 2)𝑌(𝑠) − (𝑠 2 − 2𝑠 − 1) ⋅ 0 − (𝑠 − 2) ⋅ 1 − (1) ⋅ 0 = 0
⇒ (𝑠 3 − 2𝑠 2 − 𝑠 + 2)𝑌(𝑠) − (𝑠 − 2) = 0
⇒ (𝑠 3 − 2𝑠 2 − 𝑠 + 2)𝑌(𝑠) = (𝑠 − 2)
(𝑠 − 2)
⇒ 𝑌(𝑠) =
(𝑠 3 − 2𝑠 2 − 𝑠 + 2)
𝑠−2
⇒ 𝑌(𝑠) =
(𝑠 + 1)(𝑠 − 1)(𝑠 − 2)
1
⇒ 𝑌(𝑠) =
(𝑠 + 1)(𝑠 − 1)
𝟏 𝟏 𝟏
⇒ 𝒀(𝒔) = − ( − )
𝟐 𝒔+𝟏 𝒔−𝟏

1 1 1
𝕃−1 {𝑌(𝑠)}(𝑡) = − 𝕃−1 { − } (𝑡)
2 𝑠+1 𝑠−1
1 −1 1 1 1
⇒ 𝕃−1 {𝑌(𝑠)}(𝑡) = 𝕃 { } (𝑡) + 𝕃−1 { } (𝑡)
2 𝑠+1 2 𝑠−1

𝟏 𝟏
𝒚(𝒕) = − 𝒆−𝒕 + 𝒆𝒕 = 𝐬𝐢𝐧𝐡(𝒕)
𝟐 𝟐
𝟑
2. Solve the differential equation 𝒚′′ + 𝟐 𝒚′ − 𝒚 = 𝒕𝒆−𝟐𝒕 via Laplace transform where 𝒚(𝟎) = 𝟎 and 𝒚′ (𝟎) = −𝟐.

3
𝕃 [𝑦 ′′ + 𝑦 ′ − 𝑦] (𝑠) = 𝕃[𝑡𝑒 −2𝑡 ](𝑠)
2
3
⇒ 𝕃[𝑦 ′′ ](𝑠) + 𝕃[𝑦 ′ ](𝑠) − 𝕃[𝑦](𝑠) = 𝕃[𝑡𝑒 −2𝑡 ](𝑠)
2
3 1
⇒ 𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) + (𝑠𝑌(𝑠) − 𝑦(0)) − 𝑌(𝑠) =
2 (𝑠 + 2)2
3 3 1
⇒ (𝑠 2 + 𝑠 − 1) 𝑌(𝑠) − 𝑠 ⋅ 0 − (−2) + ⋅ 0 =
2 2 (𝑠 + 2)2
1
⇒ (𝑠 − 0.5)(𝑠 + 2)𝑌(𝑠) + 2 =
(𝑠 + 2)2
1 2 1 − 2(𝑠 + 2)2 −2𝑠 2 − 8𝑠 − 7
⇒ 𝑌(𝑠) = − = =
(𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2) (𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3
𝐴 𝐵 𝐶 𝐷 −2𝑠 2 − 8𝑠 − 7
⇒ 𝑌(𝑠) = + + + =
𝑠 − 0.5 𝑠 + 2 (𝑠 + 2)2 (𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3
𝐴(𝑠 + 2)3 𝐵(𝑠 − 0.5)(𝑠 + 2)2 𝐶(𝑠 − 0.5)(𝑠 + 2) 𝐷(𝑠 − 0.5) −2𝑠 2 − 8𝑠 − 7
⇒ 𝑌(𝑠) = + + + =
(𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3 (𝑠 − 0.5)(𝑠 + 2)3

7 3 1
⇒ (𝐴 + 𝐵)𝑠 3 + (6𝐴 + 𝐵 + 𝐶) 𝑠 2 + (12𝐴 + 2𝐵 + 𝐶 + 𝐷) 𝑠 + (8𝐴 − 2𝐵 − 𝐶 − 𝐷) = −2𝑠 2 − 8𝑠 − 7
2 2 2
⇒ 𝐴 + 𝐵 = 0 ⇒ 𝑩 = −𝑨
7 7 5 𝟓
⇒ 6𝐴 + 𝐵 + 𝐶 = −2 ⇒ 6𝐴 − 𝐴 + 𝐶 = −2 ⇒ 𝐴 + 𝐶 = −2 ⇒ 𝑪 = −𝟐 − 𝑨
2 2 2 𝟐
3 15 25 𝟐𝟓
⇒ 12𝐴 + 2𝐵 + 𝐶 + 𝐷 = −8 ⇒ 12𝐴 − 2𝐴 − 3 − 𝐴 + 𝐷 = −8 ⇒ 𝐴 + 𝐷 = −5 ⇒ 𝑫 = −𝟓 − 𝑨
2 4 4 𝟒
1 5 5 25 125 23 𝟗𝟐
⇒ 8𝐴 − 2𝐵 − 𝐶 − 𝐷 = −7 ⇒ 8𝐴 + 2𝐴 + 2 + 𝐴 + + 𝐴 = −7 ⇒ 𝐴=− ⇒ 𝑨=−
2 2 2 8 8 2 𝟏𝟐𝟓
𝟗𝟐 𝟗𝟐 𝟒 𝟐
⇒ 𝑨=− , 𝑩= , 𝑪=− , 𝑫=−
𝟏𝟐𝟓 𝟏𝟐𝟓 𝟐𝟓 𝟓

𝟗𝟐 𝟗𝟐 𝟒 𝟐
⇒ 𝒀(𝒔) = − + − 𝟐
−
𝟏𝟐𝟓(𝒔 − 𝟎. 𝟓) 𝟏𝟐𝟓(𝒔 + 𝟐) 𝟐𝟓(𝒔 + 𝟐) 𝟓(𝒔 + 𝟐)𝟑

92 −1 1 92 −1 1 4 1 2 2
⇒ 𝑦(𝑡) = 𝕃−1 [𝑌(𝑠)](𝑡) = − 𝕃 [ ] (𝑡) + 𝕃 [ ] (𝑡) − 𝕃−1 [ 2
] (𝑡) − 𝕃−1 [ ] (𝑡)
125 𝑠 − 0.5 125 𝑠+2 25 (𝑠 + 2) 5 (𝑠 + 2)3

𝟗𝟐 𝟏𝒕 𝟗𝟐 −𝟐𝒕 𝟒 −𝟐𝒕 𝟐 𝟐 −𝟐𝒕


⇒ 𝒚(𝒕) = − 𝒆𝟐 + 𝒆 − 𝒕𝒆 − 𝒕 𝒆
𝟏𝟐𝟓 𝟏𝟐𝟓 𝟐𝟓 𝟓
3. Solve the differential equation 𝒚′′ − 𝒚′ = 𝐜𝐨𝐬(𝟐𝒕) + 𝐜𝐨𝐬(𝟐𝒕 − 𝟔𝝅) 𝒖(𝒕 − 𝟑𝝅) via Laplace transform where
𝒚(𝟎) = −𝟒 and 𝒚′ (𝟎) = 𝟎.

First, the given equation can be rewritten as follows.


𝑦 ′′ − 𝑦 ′ = cos(2𝑡) + cos(2(𝑡 − 3𝜋)) 𝑢(𝑡 − 3𝜋)

Then, the Laplace transform of the equation can be found below.


𝕃{𝑦 ′′ − 𝑦 ′ }(𝑠) = 𝕃{cos(2𝑡) + cos(2(𝑡 − 3𝜋)) 𝑢(𝑡 − 3𝜋)}(𝑠)
⇒ 𝕃{𝑦 ′′ − 𝑦 ′ }(𝑠) = 𝕃{cos(2𝑡)}(𝑠) + 𝕃{cos(2(𝑡 − 3𝜋)) 𝑢(𝑡 − 3𝜋)}(𝑠)
𝑠 𝑒 −3𝜋𝑠 𝑠
⇒ (𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0)) − (𝑠𝑌(𝑠) − 𝑦(0)) = +
𝑠 2 + 22 𝑠 2 + 22
𝑠 𝑒 −3𝜋𝑠 𝑠
⇒ 𝑠 2 𝑌(𝑠) − 𝑠(−4) − 0 − 𝑠𝑌(𝑠) + (−4) = +
𝑠2 + 4 𝑠2 + 4
𝑠 𝑒 −3𝜋𝑠 𝑠 𝑠 + 𝑒 −3𝜋𝑠 𝑠
⇒ (𝑠 2 − 𝑠)𝑌(𝑠) = + − 4𝑠 + 4 ⇒ 𝑠(𝑠 − 1)𝑌(𝑠) = − 4(𝑠 − 1)
𝑠2 + 4 𝑠2 + 4 𝑠2 + 4
(1 + 𝑒 −3𝜋𝑠 )𝑠 4(𝑠 − 1) 𝟏 + 𝒆−𝟑𝝅𝒔 𝟒
⇒ 𝑌(𝑠) = 2
− ⇒ 𝒀(𝒔) = 𝟐
−
𝑠(𝑠 − 1)(𝑠 + 4) 𝑠(𝑠 − 1) (𝒔 − 𝟏)(𝒔 + 𝟒) 𝒔

The partial fractions for 𝑌(𝑠) can be obtained as follows.


1 1
𝑌(𝑠) = (1 + 𝑒 −3𝜋𝑠 ) 2
−4⋅
(𝑠 − 1)(𝑠 + 4) 𝑠
1 1 𝑠+1 1
⇒ 𝑌(𝑠) = (1 + 𝑒 −3𝜋𝑠 ) ⋅ ⋅( − 2 )−4⋅
5 𝑠−1 𝑠 +4 𝑠
𝟏 𝟏 𝒔 𝟐 𝒆−𝟑𝝅𝒔 𝟏 𝒔 𝟐 𝟏
⇒ 𝒀(𝒔) = ⋅ (𝟐 ⋅ −𝟐⋅ 𝟐 − 𝟐 )+ ⋅ (𝟐 ⋅ −𝟐⋅ 𝟐 − 𝟐 )−𝟒⋅
𝟏𝟎 𝒔−𝟏 𝒔 +𝟒 𝒔 +𝟒 𝟏𝟎 𝒔−𝟏 𝒔 +𝟒 𝒔 +𝟒 𝒔

Finally, 𝑦(𝑡) is obtained by inverse Laplace transform of 𝑌(𝑠).


1 −1 1 𝑠 2 1 −1 −3𝜋𝑠 1 𝑠 2 1
𝕃−1 {𝑌(𝑠)}(𝑠) = 𝕃 {2 ⋅ −2⋅ 2 − } (𝑠) + 𝕃 {𝑒 (2 ⋅ −2⋅ 2 − )} (𝑠) − 4𝕃−1 { } (𝑠)
10 𝑠−1 𝑠 + 4 𝑠2 + 4 10 𝑠−1 𝑠 + 4 𝑠2 + 4 𝑠
𝟏 𝟏 𝟏 𝟏 𝟏 𝟏
𝒚(𝒕) = ( 𝒆𝒕 − 𝐜𝐨𝐬(𝟐𝒕) − 𝐬𝐢𝐧(𝟐𝒕) − 𝟒) 𝒖(𝒕) + ( 𝒆𝒕−𝟑𝝅 − 𝐜𝐨𝐬(𝟐𝒕 − 𝟔𝝅) − 𝐬𝐢𝐧(𝟐𝒕 − 𝟔𝝅)) 𝒖(𝒕 − 𝟑𝝅)
𝟓 𝟓 𝟏𝟎 𝟓 𝟓 𝟏𝟎
4. Solve the differential equation 𝒚′′ + 𝟑𝒚′ + 𝟐𝒚 = 𝒈(𝒕) where 𝒚(𝟎) = 𝟎, 𝒚′ (𝟎) = −𝟐 and 𝒈(𝒕) given below.
𝟐, 𝟎<𝒕<𝟔
𝒈(𝒕) = {𝒕, 𝟔 ≤ 𝒕 < 𝟏𝟎
𝟒, 𝒕 ≥ 𝟏𝟎

First, 𝑔(𝑡) can be expressed in terms of unit step functions as follows.


𝑔(𝑡) = 2 + (−2 + 𝑡)𝑢(𝑡 − 6) + (−𝑡 + 4)𝑢(𝑡 − 10)
⇒ 𝑔(𝑡) = 2𝑢(𝑡) + (𝑡 − 2)𝑢(𝑡 − 6) + (4 − 𝑡)𝑢(𝑡 − 10)
⇒ 𝑔(𝑡) = 2𝑢(𝑡) + (𝑡 − 6 + 4)𝑢(𝑡 − 6) + (−(𝑡 − 10) − 6)𝑢(𝑡 − 10)
⇒ 𝑔(𝑡) = 2𝑢(𝑡) + (𝑡 − 6)𝑢(𝑡 − 6) + 4𝑢(𝑡 − 6) − (𝑡 − 10)𝑢(𝑡 − 10) − 6𝑢(𝑡 − 10)

The given equation 𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 𝑔(𝑡) can be represented as below.


𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 2𝑢(𝑡) + (𝑡 − 6)𝑢(𝑡 − 6) + 4𝑢(𝑡 − 6) − (𝑡 − 10)𝑢(𝑡 − 10) − 6𝑢(𝑡 − 10)

Then, the Laplace transform of the equation is found as follows.


𝕃{𝑦 ′′ + 3𝑦 ′ + 2𝑦}(𝑠) = 𝕃{2𝑢(𝑡) + (𝑡 − 6)𝑢(𝑡 − 6) + 4𝑢(𝑡 − 6) − (𝑡 − 10)𝑢(𝑡 − 10) − 6𝑢(𝑡 − 10)}(𝑠)
⇒ 𝕃{𝑦 ′′ + 3𝑦 ′ + 2𝑦}(𝑠) = 𝕃{2𝑢(𝑡)}(𝑠) + 𝕃{(𝑡 − 6)𝑢(𝑡 − 6)}(𝑠) + 𝕃{4𝑢(𝑡 − 6)}(𝑠) − 𝕃{(𝑡 − 10)𝑢(𝑡 − 10)}(𝑠) − 𝕃{6𝑢(𝑡 − 10)}(𝑠)

2 𝑒 −6𝑠 4𝑒 −6𝑠 𝑒 −10𝑠 6𝑒 −10𝑠


⇒ (𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0)) + 3(𝑠𝑌(𝑠) − 𝑦(0)) + 2(𝑌(𝑠)) = + 2 + − 2 −
𝑠 𝑠 𝑠 𝑠 𝑠
2 𝑒 −6𝑠 4𝑒 −6𝑠 𝑒 −10𝑠 6𝑒 −10𝑠
⇒ 𝑠 2 𝑌(𝑠) − 𝑠 ⋅ 0 − (−2) + 3(𝑠𝑌(𝑠) − 0) + 2𝑌(𝑠) = + 2 + − 2 −
𝑠 𝑠 𝑠 𝑠 𝑠
2 + 4𝑒 −6𝑠 − 6𝑒 −10𝑠 𝑒 −6𝑠 − 𝑒 −10𝑠
⇒ (𝑠 2 + 3𝑠 + 2)𝑌(𝑠) = + −2
𝑠 𝑠2
2 + 4𝑒 −6𝑠 − 6𝑒 −10𝑠 𝑒 −6𝑠 − 𝑒 −10𝑠
⇒ (𝑠 + 1)(𝑠 + 2)𝑌(𝑠) = + −2
𝑠 𝑠2
𝟐 + 𝟒𝒆−𝟔𝒔 − 𝟔𝒆−𝟏𝟎𝒔 𝒆−𝟔𝒔 − 𝒆−𝟏𝟎𝒔 𝟐
⇒ 𝒀(𝒔) = + 𝟐 −
𝒔(𝒔 + 𝟏)(𝒔 + 𝟐) 𝒔 (𝒔 + 𝟏)(𝒔 + 𝟐) (𝒔 + 𝟏)(𝒔 + 𝟐)

The partial fractions for 𝑌(𝑠) can be obtained as follows.


1 1 1
⇒ 𝑌(𝑠) = (2 + 4𝑒 −6𝑠 − 6𝑒 −10𝑠 ) + (𝑒 −6𝑠 − 𝑒 −10𝑠 ) 2 −2
𝑠(𝑠 + 1)(𝑠 + 2) 𝑠 (𝑠 + 1)(𝑠 + 2) (𝑠 + 1)(𝑠 + 2)
⇒ 𝑌(𝑠) = (2 + 4𝑒 −6𝑠 − 6𝑒 −10𝑠 )𝑭𝟏 (𝒔) + (𝑒 −6𝑠 − 𝑒 −10𝑠 )𝑭𝟐 (𝒔) − 2𝑭𝟑 (𝒔)
⇒ 𝒀(𝒔) = 2𝑭𝟏 (𝒔) + 𝑒 −6𝑠 (4𝑭𝟏 (𝒔) + 𝑭𝟐 (𝒔)) − 𝑒 −10𝑠 (6𝑭𝟏 (𝒔) + 𝑭𝟐 (𝒔)) − 2𝑭𝟑 (𝒔)
𝐹1 (𝑠), 𝐹2 (𝑠) and 𝐹3 (𝑠) are obtained below separately.
𝐴1 𝐵1 𝐶1 1 𝟏 𝟏 𝟏 𝟏 𝟏
𝐹1 (𝑠) = + + = ⇒ 𝑭𝟏 (𝒔) = ⋅ − + ⋅
𝑠 𝑠 + 1 𝑠 + 2 𝑠(𝑠 + 1)(𝑠 + 2) 𝟐 𝒔 𝒔+𝟏 𝟐 𝒔+𝟐
𝐴2 𝐵2 𝐶2 𝐷2 1 𝟑 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏
𝐹2 (𝑠) = + 2+ + = 2 ⇒ 𝑭𝟐 (𝒔) = − ⋅ + ⋅ 𝟐 + − ⋅
𝑠 𝑠 𝑠 + 1 𝑠 + 2 𝑠 (𝑠 + 1)(𝑠 + 2) 𝟒 𝒔 𝟐 𝒔 𝒔+𝟏 𝟒 𝒔+𝟐
𝐴3 𝐵3 1 𝟏 𝟏
𝐹3 (𝑠) = + = ⇒ 𝑭𝟑 (𝒔) = −
𝑠 + 1 𝑠 + 2 (𝑠 + 1)(𝑠 + 2) 𝒔+𝟏 𝒔+𝟐

Decomposed 𝐹1 (𝑠), 𝐹2 (𝑠) and 𝐹3 (𝑠) terms can be plugged in 𝑌(𝑠) equation.
𝑌(𝑠) = 2𝐹1 (𝑠) + 𝑒 −6𝑠 (4𝐹1 (𝑠) + 𝐹2 (𝑠)) − 𝑒 −10𝑠 (6𝐹1 (𝑠) + 𝐹2 (𝑠)) − 2𝐹3 (𝑠)
1 1 1
⇒ 𝑌(𝑠) = −2⋅ +
𝑠 𝑠+1 𝑠+2
1 1 1 3 1 1 1 1 1 1
+𝑒 −6𝑠 [(2 ⋅ + 4 ⋅ +2⋅ ) + (− ⋅ + ⋅ 2 + − ⋅ )]
𝑠 𝑠+1 𝑠+2 4 𝑠 2 𝑠 𝑠+1 4 𝑠+2
1 1 1 3 1 1 1 1 1 1
−𝑒 −10𝑠 [(3 ⋅ + 6 ⋅ +3⋅ ) + (− ⋅ + ⋅ 2 + − ⋅ )]
𝑠 𝑠+1 𝑠+2 4 𝑠 2 𝑠 𝑠+1 4 𝑠+2
1 1
−2 ⋅ +2⋅
𝑠+1 𝑠+2
𝟏 𝟏 𝟏 𝟓 𝟏 𝟏 𝟏 𝟏 𝟕 𝟏
⇒ 𝒀(𝒔) = − 𝟒 ⋅ +𝟑⋅ + 𝒆−𝟔𝒔 [ ⋅ + ⋅ 𝟐 + 𝟓 ⋅ + ⋅ ]
𝒔 𝒔+𝟏 𝒔+𝟐 𝟒 𝒔 𝟐 𝒔 𝒔+𝟏 𝟒 𝒔+𝟐
𝟗 𝟏 𝟏 𝟏 𝟏 𝟏𝟏 𝟏
− 𝒆−𝟏𝟎𝒔 [ ⋅ + ⋅ 𝟐 + 𝟕 ⋅ + ⋅ ]
𝟒 𝒔 𝟐 𝒔 𝒔+𝟏 𝟒 𝒔+𝟐

Finally, 𝑦(𝑡) is obtained by inverse Laplace transform of 𝑌(𝑠).


1 1 1 5 1 1 1 1 7 1
𝕃−1 {𝑌(𝑠)} = 𝕃−1 { − 4 ⋅ +3⋅ } (𝑠) + 𝕃−1 {𝑒 −6𝑠 [ ⋅ + ⋅ 2 + 5 ⋅ + ⋅ ]} (𝑠)
𝑠 𝑠+1 𝑠+2 4 𝑠 2 𝑠 𝑠+1 4 𝑠+2
9 1 1 1 1 11 1
−𝕃−1 {𝑒 −10𝑠 [ ⋅ + ⋅ 2 + 7 ⋅ + ⋅ ]} (𝑠)
4 𝑠 2 𝑠 𝑠+1 4 𝑠+2
𝟓 𝟏 𝟕
𝒚(𝒕) = [𝟏 − 𝟒𝒆−𝒕 + 𝟑𝒆−𝟐𝒕 ]𝒖(𝒕) + [ + (𝒕 − 𝟔) + 𝟓𝒆−(𝒕−𝟔) + 𝒆−𝟐(𝒕−𝟔) ] 𝒖(𝒕 − 𝟔)
𝟒 𝟐 𝟒
𝟗 𝟏 𝟏𝟏
+ [ + (𝒕 − 𝟏𝟎) + 𝟕𝒆−(𝒕−𝟏𝟎) + 𝒆−𝟐(𝒕−𝟏𝟎) ] 𝒖(𝒕 − 𝟏𝟎)
𝟒 𝟐 𝟒

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