Question

By using $L2$, verify $L7$and$L8$ in the Laplace transform table.

Short answer

The final values are $L\left(\frac{e^{-at}-e^{-bt}}{b-a}\right)=\frac{1}{(p+a)(p+b)}$and $L\left(\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}\right)=\frac{p}{(p+a)(p+b)}$.

Step-by-step solution

Given information

From, Property ($L2)$of Laplace transforms

$$\begin{gathered} L\left(e^{-at}\right)=\frac{1}{(p+a)}\dots ..\left(1\right) \\ L\left(e^{-bt}\right)=\frac{1}{(p+b)}\dots ..\left(2\right) \end{gathered}$$

Differential equation

An equation that contains the variables and the derivatives is called a differential equation.

Solve for L7

Subtracting equation (2) from (1) as,

$$\begin{gathered} L\left(e^{-at}-\right.\left.e^{-bt}\right)=\frac{1}{(p+a)}-\frac{1}{(p+b)} \\ =\frac{p+b-(p+a)}{(p+a)(p+b)} \\ =\frac{p+b-p-a}{(p+a)(p+b)} \\ =\frac{b-a}{(p+a)(p+b)} \end{gathered}$$

Dividing both sides by $(b-a)$as,

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

Hence, to verify the expression$L\left(\frac{e^{-at}-e^{-bt}}{b-a}\right)=\frac{1}{(p+a)(p+b)}$

Solve for L8

Multiplying equation (1) with and (2) by $b$, and then subtracting equation (2) from (1) as,

$$\begin{gathered} L\left(a{e}^{-at}-b{e}^{-bt}\right)=\frac{a}{(p+a)}-\frac{a}{(p+b)} \\ =\frac{a(p+b)+b(p+a)}{(p+a)(p+b)} \\ =\frac{ap+ab-bp-ab}{(p+a)(p+b)} \\ =\frac{p(a-b)}{(p+a)(p+b)} \end{gathered}$$

Dividing both sides by role="math" localid="1664285895567" $(a-b)$as,

$$\begin{gathered} L\left(\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}\right)=\frac{p(a-b)}{(p+a)(p+b)(a-b)} \\ L\left(\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}\right)=\frac{p}{(p+a)(p+b)} \end{gathered}$$

Hence, the expression is $L\left(\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}\right)=\frac{p}{(p+a)(p+b)}$verified.