Obtain the Laplace transform of the function: 5e-2t + 3 _ 2cos 2t
The Correct Answer and Explanation is:
To obtain the Laplace transform of the function
f(t)=5e−2t+3−2cos(2t),f(t) = 5e^{-2t} + 3 – 2\cos(2t),f(t)=5e−2t+3−2cos(2t),
we will apply the basic Laplace transform rules to each term individually.
Step-by-step Laplace Transform:
- Laplace Transform of 5e−2t5e^{-2t}5e−2t: L{e−at}=1s+a,soL{5e−2t}=5s+2\mathcal{L}\{e^{-at}\} = \frac{1}{s + a}, \quad \text{so} \quad \mathcal{L}\{5e^{-2t}\} = \frac{5}{s + 2}L{e−at}=s+a1,soL{5e−2t}=s+25
- Laplace Transform of constant 333: L{3}=3s\mathcal{L}\{3\} = \frac{3}{s}L{3}=s3
- Laplace Transform of −2cos(2t)-2\cos(2t)−2cos(2t): L{cos(at)}=ss2+a2,soL{−2cos(2t)}=−2⋅ss2+4\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}, \quad \text{so} \quad \mathcal{L}\{-2\cos(2t)\} = -2 \cdot \frac{s}{s^2 + 4}L{cos(at)}=s2+a2s,soL{−2cos(2t)}=−2⋅s2+4s
Final Laplace Transform:
Now, combine all the individual transforms: L{f(t)}=5s+2+3s−2ss2+4\mathcal{L}\{f(t)\} = \frac{5}{s + 2} + \frac{3}{s} – \frac{2s}{s^2 + 4}L{f(t)}=s+25+s3−s2+42s
Explanation
The Laplace transform is a powerful integral transform used to convert time-domain functions into s-domain expressions. This is especially useful in solving linear differential equations and analyzing systems in engineering and physics.
The given function is: f(t)=5e−2t+3−2cos(2t)f(t) = 5e^{-2t} + 3 – 2\cos(2t)f(t)=5e−2t+3−2cos(2t)
We break this down into simpler parts and apply known Laplace transform formulas.
First, for exponential functions like e−ate^{-at}e−at, the Laplace transform is: L{e−at}=1s+a\mathcal{L}\{e^{-at}\} = \frac{1}{s + a}L{e−at}=s+a1
Multiplying by a constant (in this case 5) simply scales the result. So 5e−2t5e^{-2t}5e−2t becomes 5s+2\frac{5}{s + 2}s+25.
Next, for constants such as 3, the Laplace transform is: L{c}=cs\mathcal{L}\{c\} = \frac{c}{s}L{c}=sc
Hence, the transform of 3 is 3s\frac{3}{s}s3.
For trigonometric functions, the Laplace transform of cosine is: L{cos(at)}=ss2+a2\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}L{cos(at)}=s2+a2s
In this case, cos(2t)\cos(2t)cos(2t) transforms to ss2+4\frac{s}{s^2 + 4}s2+4s, and multiplying by -2 gives −2ss2+4-\frac{2s}{s^2 + 4}−s2+42s.
Adding all parts together, we get the total Laplace transform: 5s+2+3s−2ss2+4\frac{5}{s + 2} + \frac{3}{s} – \frac{2s}{s^2 + 4}s+25+s3−s2+42s
This compact s-domain expression can now be used to analyze systems or solve differential equations more easily than working in the time domain.
