Solve the differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}\mathbf{y}\mathbf{=}\mathbf{f}\left(t\right)$, here

$\mathbf{f}\left(t\right)\mathbf{=}\{\begin{array}{l}0,t<0,\\ 1,t>0,\end{array}{y}_0=y{\text{'}}_0=0.$

Hint: Use the convolution integral as in the example:

The differential equation $y\text{'}\text{'}-a^{2}y=f\left(t\right)$, here

$f\left(t\right)=\left\{\begin{array}{l}0,t<0,\\ 1,t>0,\end{array}\right.and{y}_0=y_0=0.$

Laplace transform is the integral transform of the given derivative function with real variable t to convert into a complex function

Given a second order of a differential order, with the following form

$Ay\text{'}\text{'}+By\text{'}+Cy=f\left(t\right)$ …………. (1)

Using Laplace transform and knowing that the Laplace transform of the derivatives of y is given by

$$\begin{gathered} L\left(y\text{'}\right)=pY-y_0 \\ L\left(y\text{'}\text{'}\right)=p^{2}Y-p{y}_0-y{\text{'}}_0 \end{gathered}$$

And thus take the Laplace inverse of equation (1),

$A\left(p^{2}Y-p{y}_0-y{\text{'}}_0\right)+B\left(pY-y_0\right)+CY=L^{-1}\left(f\left(t\right)\right)$

And, assume the zero boundary values,

$$\begin{gathered} A{p}^{2}Y+BpY+CY=L\left(f\left(t\right)\right) \\ Y\left(A{p}^2+Bp+C\right)=L\left(f\left(t\right)\right) \end{gathered}$$

Represent the Laplace of the function y, as

$Y=\frac{1}{\underset{partI}{\underbrace{\left(A{p}^2+Bp+C\right)}}}L\left(f\left(t\right)\right)$………….(2)

Which can be regarded as the two functions, represent part 1 of the above equation as a Laplace of some function, i.e., there must exist some function g(t) , such that the Laplace transform of this function would yield part 1, such that

Y=L(g(t))L(f(t))

here,

$L\left(g\left(t\right)\right)=\frac{1}{A{p}^{2}Bp+C}$

And thus, the Laplace of the function y can be regarded as a product of the Laplace of two function, such that

Y=G(p)F(p)

Find the Laplace inverse of the product of these two functions, such that

$y=L^{-1}\left(Y\right)=L^{-1}\left(G\left(p\right)F\left(p\right)\right)$

And the product of two Laplace transform of two function is the Laplace of the something called the convolution "symbolized by asterisk" of function g(t) and f(t) , such that

$$\begin{gathered} G\left(p\right)F\left(p\right)=L\left[{\int }_0^{t}g\left(t-\tau \right)f\left(\tau \right)d\tau \right] \\ =L\left(g*f\right) \end{gathered}$$

And, thus the Laplace inverse transform for the product of the function G(p) and F(p) which is equal to y "the solution of the differential equation" is equal to the convolution of the function g(t) and function f(t) , such that

role="math" localid="1659341414992" $$\begin{gathered} L^{-1}\left(G\left(p\right)F\left(p\right)\right)=L^{-1}\left(L\left[{\int }_0^{t}g\left(t-\tau \right)f\left(t\right)d\tau \right]\right) \\ g*f={\int }_0^{t}g\left(t-\tau \right)f\left(\tau \right)d\tau ......\left(3\right) \\ =y \end{gathered}$$

And thus if expression of Y as in equation (2), and the function f(t) is known and by find the function g(t) which would yield part 1 ,find the convolution of the two function g(t)

The function f(t) which is defined by equation (3)& (2) find the convolution find the solution of the differential equation y.

It is given that the differential equation is

$y\text{'}\text{'}-a^{2}y=f\left(t\right)$

Take, the Laplace inverse of the given differential equation,

$\left(p^{2}Y-p{y}_0-y{\text{'}}_0\right)-a^2\left(Y\right)=F\left(p\right)$

And it is given that boundary values "initial values", are zeros and hence substitute

$p^{2}Y-a^{2}Y=F\left(p\right)$

Factorize Y,

$Y\left(p^2-a^2\right)=F\left(p\right)$

Thus,

$Y=\frac{1}{p^2-a^2}$(p)

This may be considered as the product of two function G(p) and F(p), here

$G\left(p\right)=\frac{1}{p^2-a^2}$

Thus, Y is given by

Y=G(p)F(p)

And hence find the solution of this differential equation find the convolution integral of the two functions g(t) and f|(t) , i.e.

y=g*f=f*g

The function f(t), is given by

$f\left(t\right)=\left\{\begin{array}{l}0,fort<0\\ 1,fort>0\end{array}\right.$

And, from Laplace transformations table, use identity L.9 to find the Laplace inverse of the function G(p), where it would be given by

$$\begin{gathered} g\left(t\right)=L^{-1}\left(\frac{1}{p^2-a^2}\right) \\ =\frac{1}{a}\sin h\left(at\right) \end{gathered}$$

And the exponential form of the function sin h(at) , is given by

$\sin h\left(at\right)=\frac{e^{at}-e^{-at}}{2}$

Thus, the function g(t) , is given by

$g\left(t\right)=\frac{1}{a}\frac{e^{at}-e^{-at}}{2}$

Find the convolution of both function which is the solution of our differential equation, and since the function f(t) is piece-wise, then have two solution for the given differential equation, first find the convolution integral of the two function

y=g*f

And, due to symmetry of the convolution integral,

f*g=g*f

The difference it would make, is that instead of find the convolution integral of g and f ,evaluate the convolution integral of f and g, as it would be easier to evaluate where the convolution integral is defined as follows

$g*f={\int }_0^{t}g\left(t-\tau \right)f\left(\tau \right)d\tau$

And the convolution of f and g , is

$f*g={\int }_0^{t}f\left(t-\tau \right)g\left(\tau \right)d\tau$

And,find the function$f\left(t-\tau \right)$

Replace every variable t in the functionby$t-\tau$ , and since this function is not varying with t "in other words as t varies the function is a constant", then

$f\left(t-\tau \right)=f\left(\tau \right)$

At the other hand, the function g(t) itself is verify with t , and hence it would be easier to find the function $g\left(\tau \right)$as well as integrate

Calculate the convolution of f and g , thus

$$\begin{gathered} y=f*g \\ ={\int }_0^t\left(t-\tau \right)g\left(\tau \right)d\tau \end{gathered}$$

Now, two cases either$t>0$ or$t<0$ , in the latter case

$$\begin{gathered} ={\int }_0^{t}0\times \left(\frac{e^{a\tau }-e^{-a\tau }}{2a}\right)d\tau \\ =0 \end{gathered}$$

And, in the former case

$={\int }_0^{t}1\times \left(\frac{e^{a\tau }-e^{-a\tau }}{2a}\right)d\tau$

Separate, the integral

$$\begin{gathered} ={\int }_0^t\frac{e^{a\tau }}{2a}d\tau -{\int }_0^t\frac{e^{-a\tau }}{2a}d\tau \\ =({\overline{)\frac{e^{a\tau }}{2a^2}}}_0^t-({\overline{)\frac{e^{-a\tau }}{-2a^2}}}_0^t \\ =\frac{e^{a\tau }}{2a^2}-\frac{1}{2a^2}+\frac{e^{-a\tau }}{2a^2}-\frac{1}{2a^2} \end{gathered}$$

Solve further

$$\begin{gathered} =\left(\frac{1}{a^2}\frac{e^{at}+e^{-at}}{2}\right)-\frac{2}{2a^2} \\ =\left(\frac{1}{a^2}\frac{e^{at}+e^{-at}}{2}\right)-\frac{1}{2a^2} \end{gathered}$$

The exponential form of the cosine function to be

$\cos h\left(x=\frac{e^x+e^{-x}}{2}\right)$

Thus,

$$\begin{gathered} =\left(\frac{1}{a^2}\cos h\left(at\right)\right)-\frac{1}{a^2} \\ =\frac{\left(\cos h\left(at\right)-1\right)}{a^2} \end{gathered}$$

And, hence from (a) and (b) the solution of the differential equation is

$y=\left\{\begin{array}{l}0\\ \frac{\left(\cos h\left(at\right)-1\right)}{a^2}\end{array}\right.\begin{array}{c}fort<0\\ \\ fort>0\end{array}$