Laplace transform of f(x)= xcos(ax)

In summary, when using the Laplace transform to find ^f(s) from f(x) = x cos(ax), the correct solution is ^f(s) = (s^2 - a^2) / (s^2 + a^2)^2. The incorrect solution is missing the -a^2 term in the numerator. This can be found by properly expanding the expression and simplifying it.
  • #1
jhat21
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0

Homework Statement



Laplace transform from f(x) = x cos(ax)
to ^f(s) = (s^2 - a^2) / (s^2 + a^2)^2

how do you get the -a^2 term in the numerator,
all i come up with is s^2?

Homework Equations



f(x) = x e^(ax) ----> ^f(s) = 1/(s-a)^2

2 cos(ax) = e^(aix) + e^(-aix)


The Attempt at a Solution



f(s) = integral[ x/2 (e^-(s-ai)x + e^-(s+ai)x) ]dx
= 1/2[1/(s-ai)^2 + 1/(s+ai)^2]
=1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
=1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
=(s^2 )/ (s^2 + a^2)^2

but the given solution is:
(s^2 - a^2) / (s^2 + a^2)^2

so where am i missing the -a^2 term?
 
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  • #2
=1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
=1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|wrong sign!
 

1. What is the Laplace transform of f(x)= xcos(ax)?

The Laplace transform of f(x)= xcos(ax) is given by F(s) = (s^2-a^2)/(s^2+a^2)^2.

2. How do you find the Laplace transform of f(x)= xcos(ax)?

To find the Laplace transform of f(x)= xcos(ax), you can use the formula: F(s) = ∫e^(-st)f(x)dx. Plug in the values for f(x) and solve the integral to get the final result.

3. What is the significance of Laplace transform in mathematics?

Laplace transform is an important mathematical tool in solving differential equations and analyzing systems in engineering and physics. It transforms a function in the time domain into a function in the frequency domain, making it easier to solve complex problems involving differential equations.

4. Can the Laplace transform of f(x)= xcos(ax) be used to find the inverse Laplace transform?

Yes, the Laplace transform of f(x)= xcos(ax) can be used to find the inverse Laplace transform. The inverse Laplace transform of F(s) is given by f(x) = (1/2πi)∫F(s)e^(st)ds, where the integral is taken along a vertical line in the complex plane where F(s) is analytic.

5. Is the Laplace transform of f(x)= xcos(ax) a one-to-one function?

No, the Laplace transform of f(x)= xcos(ax) is not a one-to-one function. This means that different functions in the time domain can have the same Laplace transform in the frequency domain, making it impossible to uniquely determine the original function from its Laplace transform.

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