Laplace transform calculator with steps
Laplace transform calculator does the transformation of real variable function into complex variable function. This Laplace calculator gives the result of the given function with steps.
It can also provide the differential and integral of the complex variable function.
How does this Laplace transformation calculator work?
Laplace transformation is an easy-to-use tool. you can transform any real variable function into a complex variable function by following the below steps.
- Write the real variable function into the input box.
- To use the sample examples, press the load examples
- Use the keypad icon present in the input box for writing mathematical symbols.
- Press the calculate button to get the step-by-step solution of the real variable function.
- To enter a new real variable function hit the clear button.
What is Laplace transform?
A useful method for solving various kinds of the differential equation when the initial circumstances are given, especially when the initial circumstances are zero is said to be the Laplace transform. It can be defined as a function f(t) for t>0 is defined by an improper integral such as:
\(\:ℒ\left\{f\left(t\right)\right\}=\int _0^{\infty \:}e^{-st}f\left(t\right)dt\:\)
The result of the real variable function can be written as F(s). In simple words, the Laplace of f(t) with respect to t is F(s).
\( ℒ\left\{f\left(t\right)\right\}=F\left(s\right)\)
How to find the Laplace transform of a real variable function?
Here are some examples of Laplace solved by our Laplace transform solver.
Example 1
Transform \(e^{2t}+cos\left(2t\right)\) by using Laplace transform method.
Solution
Step 1: Write the given real variable functions.
\( f\left(t\right)=e^{2t}\)
\( g\left(t\right)=cos\left(2t\right)\)
Step 2: Now use the Laplace notation.
\( ℒ\left\{f\left(t\right)+g\left(t\right)\right\}=ℒ\left\{e^{2t}+cos\left(2t\right)\right\}\)
Step 3: Now Use the linearity property of Laplace and apply the notation separately.
\( ℒ\left\{e^{2t}+cos\left(2t\right)\right\}=ℒ\left\{e^{2t}\right\}+ℒ\left\{cos\left(2t\right)\right\}\)
Step 4: Now use the Laplace table and transform the above functions.
\( ℒ\left\{e^{2t}\right\}=\frac{1}{s-2}\)
\( ℒ\left\{cos\left(2t\right)\right\}=\frac{s}{s^2+4}\)
Step 5: Now write the calculated values with the given function.
\( ℒ\left\{e^{2t}+cos\left(2t\right)\right\}=\frac{1}{s-2}+\frac{s}{s^2+4}\)
Example 2
Find the Laplace transform of \( e^{-5t}+cos^2\left(2t\right)\)
Solution
Step 1: Write the given real variable functions.
\( f\left(t\right)=e^{-5t}\)
\( g\left(t\right)=cos^2\left(2t\right)\)
Step 2: Now use the Laplace notation.
\( ℒ\left\{f\left(t\right)+g\left(t\right)\right\}=ℒ\left\{e^{-5t}+cos^2\left(2t\right)\right\}\)
Step 3: Now Use the linearity property of Laplace and apply the notation separately.
\( ℒ\left\{e^{-5t}+cos^2\left(2t\right)\right\}=ℒ\left\{e^{-5t}\right\}+ℒ\left\{cos^2\left(2t\right)\right\}\)
Step 4: Now use the Laplace table and transform the above functions.
\( ℒ\left\{e^{-5t}\right\}=\frac{1}{s+5}\)
\( ℒ\left\{cos^2\left(2t\right)\right\}=\frac{1}{s}+\frac{s}{2\left(s^2+16\right)}\)
Step 5: Now write the calculated values with the given function.
\( ℒ\left\{e^{-5t}+cos^2\left(2t\right)\right\}=\frac{1}{s+5}+\frac{1}{2s}+\frac{s}{2\left(s^2+16\right)}\)
Table of Laplace transform
Below is the table of Laplace.
f(t) | \(F\left(s\right)=L\left\{f\left(t\right)\right\}\) |
1 | \(\frac{1}{s}\) |
\(e^{at}\) | \(\frac{1}{s-a}\) |
\(t^n,n=1,2,3,...\) | \(\frac{n}{s^{n+1}}\) |
\(t^p, p>-1\) | \(\frac{Γ\left(p+1\right)}{s^{p+1}}\) |
\(\sqrt{t}\) | \(\frac{\sqrt{\pi \:}}{2s^{\frac{3}{2}}}\) |
\(sin\left(at\right)\) | \(\frac{a}{s^2+a^2}\) |
\(cos\left(at\right)\) | \(\frac{s}{s^2+a^2}\) |
\(tsin\left(at\right)\) | \(\frac{2as}{\left(s^2+a^2\right)^2}\) |
\(tcos\left(at\right)\) | \(\frac{s^2-a^2}{\left(s^2+a^2\right)^2}\) |
\(sin\left(at\right)−atcos\left(at\right)\) | \(\frac{2a^3}{\left(s^2+a^2\right)^2}\) |
\(sin\left(at\right)+atcos\left(at\right)\) | \(\frac{2as^2}{\left(s^2+a^2\right)^2}\) |
\(cos\left(at\right)−atsin\left(at\right)\) | \(\frac{s\left(s^2-a^2\right)}{\left(s^2+a^2\right)^2}\) |
\(cos\left(at\right)+atsin\left(at\right)\) | \(\frac{s\left(s^2+)3a^2\right)}{\left(s^2+a^2\right)^2}\) |
\(sin\left(at+b\right)\) | \(\frac{ssin\left(b\right)+acos\left(b\right)}{s^2+a^2}\) |
\(cos\left(at+b\right)\) | \(\frac{scos\left(b\right)−asin\left(b\right)}{s^2+a^2}\) |
\(sinh\left(at\right)\) | \(\frac{a}{s^2-a^2}\) |
\(cosh\left(at\right)\) | \(\frac{s}{s^2-a^2}\) |
\(e^{at}sin\left(bt\right)\) | \(\frac{b}{\left(s-a\right)^2+b^2}\) |
\(e^{at}cos\left(bt\right)\) | \(\frac{s-a}{\left(s-a\right)^2+b^2}\) |
\(e^{at}sinh\left(bt\right)\) | \(\frac{b}{\left(s-a\right)^2-b^2}\) |
\(e^{at}cosh\left(bt\right)\) | \(\frac{s-a}{\left(s-a\right)^2-b^2}\) |
\(t^ne^{at},\:n=1,2,3,...\) | \(\frac{n!}{\left(s-a\right)^{n+1}}\) |
\(f\left(ct\right)\) | \(\frac{1}{s}F\left(\frac{s}{c}\right)\) |
\(uc\left(t\right)=u\left(t−c\right)\) | \(e^{-cs}\) /s |
\(δ\left(t−c\right)\) | \(e^{-cs}\) |
\(u_c\left(t\right)f\left(t−c\right)\) | \(e^{-cs}F\left(s\right)\) |
\(u_c\left(t\right)g\left(t\right)\) | \(e^{-cs}L\left\{g\left(t+c\right)\right\}\) |
\(e^{ct}f\left(t\right)\) | \(F\left(s−c\right)\) |
\(t^nf\left(t\right),n=1,2,3,...\ \) | \(\left(-1\right)^nF^{\left(n\right)}\left(s\right)\) |
\(\frac{1}{t}f\left(t\right)\) | \(\int _s^{\infty }F\left(u\right)du\) |
\(\int _0^tf\left(v\right)dv\) | F(s)/s |
\(\int _0^tf\left(t−τ\right)g\left(τ\right)dτ\) | \(F\left(s\right)G\left(s\right)\) |
\(f\left(t+T\right)=f\left(t\right)\) | \(\frac{\int _0^Te^{-st}f\left(t\right)dt\:}{1-e^{-sT}}\) |
\(f'\left(t\right)\) | \(sF\left(s\right)−f\left(0\right)\) |
\(f''\left(t\right)\) | \(s^2F\left(s\right)-s f\left(0\right)−f′\left(0\right)\) |