Question

Use theorem 7.1.1 to find$L\left\{f\left(t\right)\right\},$if$f\left(t\right)=\cos 5t+\sin 2t?$

Short answer

The Laplace transform of given function is,

$L\left\{\cos 5t+\sin 2t\right\}=\frac{s}{s^2+25}+\frac{2}{s^2+4},s>a$

Step-by-step solution

Step 1:

Here we determine the Laplace transform of the given function by comparing it to general form as provided in theorem 7.1.1.

$L\left\{\sin at\right\}=\frac{a}{s^2+a^2},s>a$

$L\left\{\cos at\right\}=\frac{s}{s^2+a^2},s>a$

Step 2:

Consider the function $f\left(t\right)=\cos 5t+\sin 2t$

The objective is to find $L\left\{f\left(t\right)\right\}$using the theorem.

From the linearity property of the Laplace transform, we have

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{\cos 5t+\sin 2t\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{\cos 5t\right\}+L\left\{\sin 2t\right\} \end{gathered}$$

From the theorem7.1.1,

$L\left\{\sin at\right\}=\frac{a}{s^2+a^2},s>a$, $L\left\{\cos at\right\}=\frac{s}{s^2+a^2},s>a$Where$a=0,1,2,3,............$

$$\begin{gathered} L\left\{\cos 5t+\sin 2t\right\}=\frac{s}{s^2+5^2}+\frac{2}{s^2+2^2} \\ L\left\{\cos 5t+\sin 2t\right\}=\frac{s}{s^2+25}+\frac{2}{s^2+4},s>a \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{\cos 5t+\sin 2t\right\}=\frac{s}{s^2+25}+\frac{2}{s^2+4},s>a$