Laplace Transform

Whutever42

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Jan 19, 2018
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Currently taking a diffeq course and we learned about how to take a Laplace transform of various functions. We were told HOW to do the transform but not really WHAT the Laplace Transform is. I was wondering if someone could explain what exactly the Laplace transform is. For example why is the transform of cos(at) = s/(s2+a2)? Thank you ahead of time!
 
I am puzzled by this. If you learned "how to take a Laplace Transform" why do you not know how to find the Laplace transform of cos(at)? The Laplace transform of a function, f(x), is 0f(t)estdt\displaystyle \int_0^\infty f(t)e^{-st}dt. In particular applying that to cos(at)\displaystyle cos(at) means finding 0cos(at)estdt\displaystyle \int_0^\infty cos(at)e^{st}dt. Since you are integrating a product of two functions, that typically means "integration by parts". Here, let u(t)=cos(at)\displaystyle u(t)= cos(at) and dv=estdt\displaystyle dv= e^{-st}dt. Then du=asin(at)dt\displaystyle du= -a sin(at)dt and v=1sest\displaystyle v= -\frac{1}{s}e^{-st}. So 0cos(ax)estdt=[1scos(at)est]0as0sin(at)estdt=1sas0sin(ax)estdt\displaystyle \int_0^\infty cos(ax)e^{-st} dt= \left[\frac{1}{s}cos(at)e^{-st}\right]_0^\infty- \frac{a}{s}\int_0^\infty sin(at)e^{-st}dt= -\frac{1}{s}- \frac{a}{s}\int_0^\infty sin(ax)e^{-st}dt. Now, use integration by parts again. Let u= sin(at) and let dv=estdt\displaystyle dv= e^{-st}dt. Then du=acos(at)dt\displaystyle du= acos(at)dt and v=1sestdt\displaystyle v= -\frac{1}{s}e^{-st}dt. 0sin(at)estdt=[1ssin(at)est]0+as0cos(at)estdt=as0cos(at)estdt\displaystyle \int_0^\infty sin(at)e^{-st}dt= \left[-\frac{1}{s}sin(at)e^{-st}\right]_0^\infty+ \frac{a}{s}\int_0^\infty cos(at)e^{-st}dt= \frac{a}{s}\int_0^\infty cos(at)e^{-st}dt.

Putting that into the first integral, 0cos(t)estdt=1sa2s20cos(t)estdt\displaystyle \int_0^\infty cos(t)e^{-st}dt= \frac{1}{s}- \frac{a^2}{s^2}\int_0^\infty cos(t)e^{-st}dt. Then (1+a2s2)0cos(t)estdt=1s\displaystyle \left(1+ \frac{a^2}{s^2}\right)\int_0^\infty cos(t)e^{-st}dt= \frac{1}{s}. Dividing both sides by 1+a2s2=s2+a2s2\displaystyle 1+ \frac{a^2}{s^2}=\frac{s^2+ a^2}{s^2} gives 0cos(t)estdt=ss2+a2\displaystyle \int_0^\infty cos(t)e^{-st}dt= \frac{s}{s^2+ a^2}.

It is important for differential equations because the Laplace transform of the derivative of function f is an algebraic formula: To find 0f(t)estdt\displaystyle \int_0^\infty f'(t)e^{-st}dt again use integration by parts letting u=est\displaystyle u= e^{-st} and dv=f(t)dt\displaystyle dv= f'(t)dt so that du=sestdt\displaystyle du= -se^{-st}dt and v=f(t)\displaystyle v= f(t). So 0f(t)estdt=[f(t)est]0s0f(t)estdt=f(0)s0f(t)estdt\displaystyle \int_0^\infty f'(t)e^{-st}dt= \left[f(t)e^{-st}\right]_0^\infty- s\int_0^\infty f(t)e^{-st}dt= f(0)- s\int_0^\infty f(t)e^{-st}dt. That is, the Laplace transform of the derivative of f is f(0) minus the Laplace transform of f. A differential equation in f can be converted to an algebraic equation for the Laplace transform of f.
 
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