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Solve the following differential equations. Express the solution of the given initial value problem in terms of convolution integral. a) y"+w²y = g(t), y(0) = 0, y'(0) = 1 3.

Solve the following differential equations. Express the solution of the given initial value problem in terms of convolution integral. a) y"+w²y = g(t), y(0) = 0, y'(0) = 1 3.

Advanced Engineering Mathematics
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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**Problem 3: Differential Equations and Convolution Integrals** Solve the following differential equations. Express the solution of the given initial value problem in terms of a convolution integral. a) \( y'' + \omega^2 y = g(t), \quad y(0) = 0, \quad y'(0) = 1 \) b) \( y'' + 2y' + y = \int_{0}^{t} \tau e^{-(t-\tau)} \, d\tau, \quad y(0) = y'(0) = 0 \)
Transcribed Image Text:**Problem 3: Differential Equations and Convolution Integrals** Solve the following differential equations. Express the solution of the given initial value problem in terms of a convolution integral. a) \( y'' + \omega^2 y = g(t), \quad y(0) = 0, \quad y'(0) = 1 \) b) \( y'' + 2y' + y = \int_{0}^{t} \tau e^{-(t-\tau)} \, d\tau, \quad y(0) = y'(0) = 0 \)
### Formulas to Remember 1. **Laplace Transform:** \[ \mathcal{L}(e^{at}f(t)) = F(s-a); \] This formula shows the Laplace transform of a function multiplied by an exponential term. 2. **Inverse Laplace Transform:** \[ \mathcal{L}^{-1}(F(s-a)) = e^{at}f(t) \quad \text{where} \quad f(t) = \mathcal{L}^{-1}(F(s)) \] The inverse Laplace transform retrieves the original function with an exponential term. 3. **Laplace Transform with Unit Step Function:** \[ \mathcal{L}(f(t-c)U_c(t)) = e^{-cs}L(f(t)); \] This applies the Laplace transform to functions with delayed unit step functions. 4. **Inverse Laplace Transform with Unit Step Function:** \[ \mathcal{L}^{-1}(e^{-cs}F(s)) = f(t-c)U_c(t) \quad \text{where} \quad f(t) = \mathcal{L}^{-1}(F(s)) \] This formula helps in identifying shifted functions in time domain. 5. **Laplace of Delta Function:** \[ \mathcal{L}(\delta(t-c)) = e^{-cs}; \] It shows the Laplace transform of a shifted delta function. 6. **Product of Laplace Transforms:** \[ \mathcal{L}(f(t)g(t)) = F(s)G(s); \] The Laplace transform of a product of functions is a product of their transforms. 7. **Inverse Product of Laplace Transforms:** \[ \mathcal{L}^{-1}(F(s)G(s)) = f(t) \ast g(t) \] This is the inverse Laplace corresponding to the convolution product. 8. **Convolution in Time Domain:** \[ f(t) \ast g(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau = \int_{0}^{t} g(\tau)f(t-\tau)d\tau \] It defines the convolution of two
Transcribed Image Text:### Formulas to Remember 1. **Laplace Transform:** \[ \mathcal{L}(e^{at}f(t)) = F(s-a); \] This formula shows the Laplace transform of a function multiplied by an exponential term. 2. **Inverse Laplace Transform:** \[ \mathcal{L}^{-1}(F(s-a)) = e^{at}f(t) \quad \text{where} \quad f(t) = \mathcal{L}^{-1}(F(s)) \] The inverse Laplace transform retrieves the original function with an exponential term. 3. **Laplace Transform with Unit Step Function:** \[ \mathcal{L}(f(t-c)U_c(t)) = e^{-cs}L(f(t)); \] This applies the Laplace transform to functions with delayed unit step functions. 4. **Inverse Laplace Transform with Unit Step Function:** \[ \mathcal{L}^{-1}(e^{-cs}F(s)) = f(t-c)U_c(t) \quad \text{where} \quad f(t) = \mathcal{L}^{-1}(F(s)) \] This formula helps in identifying shifted functions in time domain. 5. **Laplace of Delta Function:** \[ \mathcal{L}(\delta(t-c)) = e^{-cs}; \] It shows the Laplace transform of a shifted delta function. 6. **Product of Laplace Transforms:** \[ \mathcal{L}(f(t)g(t)) = F(s)G(s); \] The Laplace transform of a product of functions is a product of their transforms. 7. **Inverse Product of Laplace Transforms:** \[ \mathcal{L}^{-1}(F(s)G(s)) = f(t) \ast g(t) \] This is the inverse Laplace corresponding to the convolution product. 8. **Convolution in Time Domain:** \[ f(t) \ast g(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau = \int_{0}^{t} g(\tau)f(t-\tau)d\tau \] It defines the convolution of two
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