To show that the solution $y(t)$ to the differential equation

$a y'' + b y' + c y = f(t)$

with initial conditions $y(0) = y'(0) = 0$ is given by both formulas:

$y(t) = f(0)B'(t) + f'(0)B(t) + \int_0^t B(s)f''(t - s) \, ds,$

and

$y(t) = \int_0^t B''(s) f(t - s) \, ds,$

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Step 1: Understand the properties of $B(t)$

Given that $B(t)$ is defined by

$a B''(t) + b B'(t) + c B(t) = t,$

with initial conditions $B(0) = B'(0) = 0$, this implies that $B(t)$ is the solution to the differential equation driven by the gradual input function $f(t) = t$.

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Step 2: Use convolution for the general solution

To find the solution for $y(t)$ in terms of $f(t)$, recall that the general solution to linear differential equations with zero initial conditions can be written using the Green's function approach as

$y(t) = \int_0^t G(t - s) f(s) \, ds,$

where $G(t)$ is the impulse response function.

Since the differential equation governing $B(t)$ is driven by a simple input $t$, $B(t)$ effectively behaves like a Green's function integrated against $f(t)$.

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Step 3: Derivation of the first formula

To derive

$y(t) = f(0)B'(t) + f'(0)B(t) + \int_0^t B(s) f''(t - s) \, ds,$

perform a Taylor series expansion of the input function $f(t)$ around $t = 0$:

$f(t) = f(0) + f'(0) t + \int_0^t (t - s) f''(s) \, ds.$

Applying the linearity of the differential equation, the solution for each component of this decomposition is:

1. The term proportional to $f(0)$ yields a contribution proportional to $B'(t)$ because differentiating $B(t)$ once gives the solution response to a constant input.

2. The term proportional to $f'(0)$ yields a contribution proportional to $B(t)$ since the equation $a B'' + b B' + c B = t$ is already structured for a linear input.

3. The integral term contributes the convolution of $B(s)$ with the second derivative $f''(t - s)$.

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Step 4: Derivation of the second formula

To derive

$y(t) = \int_0^t B''(s) f(t - s) \, ds,$

apply integration by parts to the integral term in the first formula:

$\int_0^t B(s) f''(t - s) \, ds.$

Integrating by parts with $u = B(s)$, $dv = f''(t - s) ds$, and carefully accounting for boundary conditions gives

$\int_0^t B(s) f''(t - s) \, ds = \int_0^t B''(s) f(t - s) \, ds - f(0) B'(t) - f'(0) B(t).$

Thus, combining these terms yields

$y(t) = f(0) B'(t) + f'(0) B(t) + \int_0^t B''(s) f(t - s) \, ds.$

This agrees with both required expressions.

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Final Answer

Thus, the solution $y(t)$ is indeed given by both forms:

$y(t) = f(0)B'(t) + f'(0)B(t) + \int_0^t B(s) f''(t - s) ds,$

and

$y(t) = \int_0^t B''(s) f(t - s) ds.$

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Would you like further details, examples, or clarifications on any step?

5 Related Questions:

1. What is the significance of using convolution in differential equations?
2. How does the Green's function approach relate to these results?
3. Can we generalize this result for non-zero initial conditions?
4. What are the practical applications of these integral solutions?
5. How does integration by parts help in simplifying convolution integrals?

Tip: The Green's function approach provides a powerful framework for solving linear differential equations with arbitrary input functions.