Consider the following initial value problem: Edit y" + 8y' + 15y = d(t – 5) + u10(t); y(0) = 0, y(0) 4 a) Find the solution y(t). y(t) 8. -3t - e 1 -3(t-5) e 1 -5(t-5) u(t) X 1 e5(t–10) 10 1 -3(t-10) e ua(t) 6. 15 where c= and d =10

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Initial Value Problem and Its Solution**

Consider the following initial value problem:

\[ y'' + 8y' + 15y = \delta(t - 5) + u_{10}(t); \quad y(0) = 0, \quad y'(0) = \frac{1}{4} \]

**a) Find the solution \( y(t) \).**

The solution to the problem is given by:

\[ y(t) = \frac{1}{8} \left( e^{-3t} - e^{-5t} \right) \]

\[ + \left( \frac{1}{2} \left( e^{-3(t-5)} - \frac{1}{2} e^{-5(t-5)} \right) \right) u_c(t) \]

\[ + \left( -\frac{1}{6} e^{-3(t-10)} + \frac{1}{10} e^{-5(t-10)} + \frac{1}{15} \right) u_d(t) \]

where \( c = 5 \) and \( d = 10 \).
Transcribed Image Text:**Initial Value Problem and Its Solution** Consider the following initial value problem: \[ y'' + 8y' + 15y = \delta(t - 5) + u_{10}(t); \quad y(0) = 0, \quad y'(0) = \frac{1}{4} \] **a) Find the solution \( y(t) \).** The solution to the problem is given by: \[ y(t) = \frac{1}{8} \left( e^{-3t} - e^{-5t} \right) \] \[ + \left( \frac{1}{2} \left( e^{-3(t-5)} - \frac{1}{2} e^{-5(t-5)} \right) \right) u_c(t) \] \[ + \left( -\frac{1}{6} e^{-3(t-10)} + \frac{1}{10} e^{-5(t-10)} + \frac{1}{15} \right) u_d(t) \] where \( c = 5 \) and \( d = 10 \).
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