Hello, could I get some help with understanding a differential equations question involving Laplace transforms, the heaviside function, and the dirac delta?My work is attached, but I'm not understanding the second part of the solving process, when applying the inverse Laplace transform to yield an answer of et if 0 <= t < 1.Thank you! Certainly! Below is the transcription of the image, formatted for an educational website. ---### Solving Differential Equations with the Laplace Transform Method#### Problem StatementGiven the differential equation:\[ y'' + 2y' - 3y = \delta(t-1) \]with initial conditions $ y(0) = 1 $ and $ y'(0) = 1 $.#### Step-by-Step Solution1. **Transform the Differential Equation:** \[ \mathcal{L}\{y''\} + 2\mathcal{L}\{y'\} - 3\mathcal{L}\{y\} = \mathcal{L}\{\delta(t-1)\} \]2. **Apply Laplace Transforms:** \[ s^2 Y(s) - s y(0) - y'(0) + 2 \left[ s Y(s) - y(0) \right] - 3 Y(s) = e^{-s} \] Plugging in the initial conditions $ y(0) = 1 $ and $ y'(0) = 1 $: \[ [s^2 Y(s) - s - 1] + 2[s Y(s) - 1] - 3 Y(s) = e^{-s} \]3. **Simplify and Combine Terms:** \[ [s^2 Y(s) - s - 1] + 2[s Y(s) - 1] - 3 Y(s) = e^{-s} \] \[ s^2 Y(s) - s - 1 + 2sY(s) - 2 - 3Y(s) = e^{-s} \] Combine: \[ (s^2 + 2s - 3) Y(s) - s - 3 = e^{-s} \]4. **Solve for $ Y(s) $:** \[ Y(s) = \frac{s}{s^2 + 2s - 3} + \frac{3}{s^2 + 2s - 3} + e^{-s} \frac{1}{s^2 + 2s - 3} \]5. **Inverse Laplace Transform:** \[ 2 L^{-1} \left\

We are using partial fraction to find inverse Laplace of given functions.

у" +2y' -зу Sh²+-1) y (o) = L {y" } + 2 α{y ¹³} −3 £ {y} = £ { $(+-1)} [S².Y(s) — 5. y/o) - y₁(6)] + 2 [S₁Y(s) - 410)] -3 [yrs] = y (s) 22+{s S [5².4₁5) -5=1] + 2[5.425)-1] −3 [yLs) = e-s S ycs) [5²+25 -3 ] −5 −3 = e-s = s 52 +25-3 + 5²+25-3 3 52 +25-3 }+36-151 f (S²+25-3) te + = 1₁ y ₁ (0) = 1 -s T 5²+25-3 e-s L {es. I (5²+25-3). 2 I t ¨ ( ²4/7 e ² + ²/ 0-²²) + 3 ( 4 ² ² -4e-²x) | H(+-1) (4e²+ - texans) _-3(+-1) -31 et 4 е че 个个 for 0² +

у" +2y' -зу Sh²+-1) y (o) = L {y" } + 2 α{y ¹³} −3 £ {y} = £ { $(+-1)} [S².Y(s) — 5. y/o) - y₁(6)] + 2 [S₁Y(s) - 410)] -3 [yrs] = y (s) 22+{s S [5².4₁5) -5=1] + 2[5.425)-1] −3 [yLs) = e-s S ycs) [5²+25 -3 ] −5 −3 = e-s = s 52 +25-3 + 5²+25-3 3 52 +25-3 }+36-151 f (S²+25-3) te + = 1₁ y ₁ (0) = 1 -s T 5²+25-3 e-s L {es. I (5²+25-3). 2 I t ¨ ( ²4/7 e ² + ²/ 0-²²) + 3 ( 4 ² ² -4e-²x) | H(+-1) (4e²+ - texans) _-3(+-1) -31 et 4 е че 个个 for 0² +

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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