In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$

Question

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$

Accepted Answer

The solution of the given initial value problem using the method of Laplace transforms is

$y (t) = 2 e^{- 2 t} - 2 e^{- 3 t} + [\frac{1}{36} + \frac{1}{6} (t - 1) - \frac{1}{4} e^{- 2 (t - 1)} + \frac{2}{9} e^{- 3 (t - 1)}] u (t - 1) - [\frac{19}{36} + \frac{1}{6} (t - 5) - \frac{7}{4} e^{- 2 (t - 5)} + \frac{11}{9} e^{- 3 (t - 5)}] u (t - 5)]$

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Fundamentals Of Differential Equations And Boundary Value Problems

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In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform:

Step 3: Take inverse Laplace transform we get

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.y''+5y'+6y=g(t):y(0)=0,y'(0)=2, where g(t)={0 , 0⩽t<1,t , 1<t<5,1 ,5<t

Step by Step Solution

Step 1: Define Laplace Transform

Step 2: Apply Laplace transform:

Step 3: Take inverse Laplace transform we get

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In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.
$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$