Since the Laplace transforms of the derivatives of $f(t)$ are polynomials in the transform parameter $s$ (see table of Laplace transforms), forming the Laplace transform of a linear differential equation with constant coefficients and initial conditions at $t=0$ yields generally a simple equation (image equation) for solving the transformed function $F(s)$. Since the initial conditions can be taken into consideration instantly, one needs not to determine the general solution of the differential equation.
For example, transforming the equation
$f^{\prime\prime}(t)+2f^{\prime}(t)+f(t)=e^{-t}\qquad(f(0)=0,\;\;f^{\prime}(0)=1)$ |
gives
$[s^{2}F(s)-sf(0)-f^{\prime}(0)]+2[sF(s)-f(0)]+F(s)=\frac{1}{s+1},$ |
i.e.
$(s^{2}+2s+1)F(s)=1+\frac{1}{s+1},$ |
whence
$F(s)=\frac{1}{(s+1)^{2}}+\frac{1}{(s+1)^{3}}.$ |
Taking the inverse Laplace transform produces the result
$f(t)\,=\,te^{-t}+\frac{t^{2}e^{-t}}{2}\;=\;\frac{e^{-t}}{2}(t^{2}+2t).$ |
Comments
to or for?
Is "to" the right preposition in the title of http://planetmath.org/encyclopedia/UsingLaplaceTransformToInitialValuePr... ?
Maybe "for" were better?
Jussi
Re: to or for?
"For" would probably be better. Mathprof's suggestion is also good.
Cam
Re: to or for?
Thanks, Cam. I think "for" is better since it yields a bit shorter title =o)
Jussi
Re: to or for? (from Europe)
My dear Cam, Jussi and Thomas,
I think the title suggested by Thomas is clearly better.
perucho
Re: to or for? (from Europe)
Dear Peruchin,
You may be right, but how do you justify it?
Jussi
Re: to or for? (from Europe)
Well my dear friend, I think that this question is not so important (like seems suggest mathcam). When you refer to a initial values by using Laplace's transform, my opinion is that Thomas suppose you're solving an ODE, hence his suggestion seems really reasonable.
Or I'm missing something?
peruchin
Re: to or for? (from Europe)
Dear Perucho,
although I love short titles, I will change the title as you and Thomas wish =o)
Regards,
Jussi