Question

Using the $\mathit{\delta }$function method, find the response (see Problem 6c) of each of the following systems to a unit impulse.

10. ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{9}\mathit{y}\mathbf{=}\mathit{\delta }\left(t-t_0\right)$

Short answer

The given function is verified

Step-by-step solution

Given information

The given expressions are $y^2-9y=\delta \left(t-t_0\right)$.

Definition of Laplace Transformation 

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t).

Verify the given function

Assuming, that the given differential equation represent the motion of a system initially at rest, i.e. the boundary conditions are zeros

$y_0=0y_0^{\text{'}}=0$

And, hence taking the La Place transformation for the given differential equation, and substituting by the boundary values, we get the following equation

$$\begin{gathered} p^{2}Y-9Y=L\left(\delta \left(t-t_0\right)\right) \\ Y\left(p^2-9\right)=L\left(\delta \left(t-t_0\right)\right) \end{gathered}$$

Where, the bracket is a difference between squares, such that

$\left(p^2-9\right)=(p-3)(p+3)$

And, hence we have

$Y(p-3)(p+3)=L\left(\delta \left(t-t_0\right)\right)$

And, thus Y is expressed as

$Y=\frac{1}{(p-3)(p+3)}L\left(\delta \left(t-t_0\right)\right)$ …………………(a)

we have obtained from equation (a), is the product of two functions H(p) and G(p), where

$$\begin{gathered} H\left(p\right)=L\left(\delta \left(t-t_0\right)\right) \\ G\left(p\right)=\frac{1}{(p-3)(p+3)} \end{gathered}$$

And, since we need to find the La Place inverse of Y, in order to find the solution to the differential equation, which would be the given by the following

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

Where, the La Place inverse of the product of two function, can be found using the convolution integral, and the convolution integral is defined as

$L^{-1}\left(G\right(p\left)H\right(p\left)\right)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau$

And, the function g(t) is the La Place inverse of the function G(p), and the function $h(t)$ is the La Place inverse of the function G(p), where we can use the identity L.7 and L.27 to find the La Place inverse of the function G(p) and H(p), thus we have

$L^{-1}\left(H\right(p\left)\right)=L^{-1}\left(L\left(\delta \left(t-t_0\right)\right)\right)L^{-1}\left(G\right(p\left)\right)=L^{-1}\left(\frac{1}{(p+3)(p-3)}\right)$

$h\left(t\right)=\delta \left(t-t_0\right)$

$g\left(t\right)=\frac{e^{3t}-e^{-3t}}{3-(-3)}$

$=\frac{e^{3t}-e^{-3t}}{6}$

And, hence knowing the function g(t) and h(t), we can find the La Place inverse of the product of the two function G(p) and H(p), using the convolution integral which is given by

$L^{-1}\left(G\right(p\left)H\right(p\left)\right)=g*h={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau$

And the function$g(t-\tau )$is obtained by replacing every variable t with$t-\tau$in function g(t), and similarly we get the function$h\left(\tau \right)$by replacing every variablein function with$\tau$, and hence we have

localid="1664349987534" $$\begin{gathered} g(t-\tau )=\frac{e^{3(t-\tau )}-e^{-3(t-\tau )}}{6} \\ h\left(\tau \right)=\delta \left(\tau -t_0\right) \end{gathered}$$

And, hence the inverse La Place of the given function using the convolution integral would be

localid="1664349994196" $y={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau ={\int }_0^t\frac{e^{3(t-\tau )}-e^{-3(t-\tau )}}{6}\delta \left(\tau -t_0\right)d\tau$

Where, the solution of such integral is given by equation (11.6), which states that an integral

localid="1664350002927" $I={\int }_0^t\varphi \left(x\right)\delta (x-a)dx$

Is given by,

localid="1664350011217" $=\varphi \left(a\right)$

Provided that, the value of the boundary value t is greater than a, else the integral is zero. In other words the integration picks up the value of a, if the range of the integral composes the value of a " a lies between 0 and t", and hence the convolution integral is

localid="1664350018830" $=\frac{e^{3\left(t-t_0\right)}-e^{-3\left(t-t_0\right)}}{6}$

Provided that, the value of t is greater than t₀, i.e t>t₀. else the integral is zero and hence the La Place inverse of the given function is