Laplace Transform ODE Solving: y'' + 2y' + y = 0 with Initial Conditions

[IAmClifford]

Solve using laplace.

The diff eq is
y'' + 2*y' + *y = 0 subject to y(0)=1 and y(pi)=0

Sorry if notation isn't the norm. y'' and y' and y are time (t) based functions.

 

[jackmell]

Can you take the Laplace transform of everything? For example, I can differentiate the transform:

$$\frac{d}{ds}\left\{F(s)=L(f)=\int_0^{\infty}e^{-st}f(s)ds\right\}$$

and obtain:

$$\frac{dF}{ds}=-\int_0^{\infty}e^{-st}}[tf(t)]dt$$

so that

$$L\left\{tf(t)\right\}=-\frac{dF}{ds}$$

and there are other formulas for the other terms as well so that when you transform everything, you'll get another DE in terms of the transform F(s) which you then solve for F(s), then invert it.

 

[JJacquelin]

The diff eq is
t*y'' + 2*y' + t*y = 0 subject to y(0)=1 and y(pi)=0

Is there no mistake in the wording of the problem ? Because solving this ODE with Laplace Transform is much more difficult than with traditional method.
Traditionally, the result is :
y(t) = (a*BesselJ[1/2 , t]+b*BesselJ[-1/2 , t] ) / Sqrt(t)
a, b = constants
With Laplace Transform method :
Notation : Laplace Transform of y(t) is g(s)
Laplace Transform of t*y(t) is -g'(s)
Laplace Transform of y'(t) is s*g(s)-y(0) = s*g(s)-1
Laplace Transform of y''(t) is s²g(s)-s*y(0)-y'(0) = s²g-s-y'(0)
Laplace Transform of t*y''(t) is -(s²g-s-y'(0) )' = -2s*g-s²g'(s)+1
Laplace Transform of t*y'' + 2*y' + t*y = 0 is :
-2s*g-s²g'(s)+1+2(s*g(s)-1)-g'(s) = 0
g'(s) = -1/(1+s²)
g(s) = -arctan(s) + Constant
So, we come across a snag : finding the Inverse Laplace Transform of arctan(s) ? Of coiurse, it should lead to the formula with the Bessel functions.

 

[JJacquelin]

I bet you that the equation is t*y'' + 2*y' + t*y = 2 cos(t)

 

[JJacquelin]

Its not, it's simply equal to 0.

OK. As a matter of fact, solving is simpler than it appears at first sight :
The key point : No need to integrate g'(s) = -1/(1+s²)
The Inverse Laplace Transform of g'(s) is -t*y(t)
The Inverse Laplace Transform of 1/(1+s²) is sin(t)
Then -t*y(t) = -sin(t)
The solution is y(t) = sin(t)/t
y(0) = limit sin(t)/t =1
y(pi) = sin(pi)/pi = 0

 

[JJacquelin]

Note that the general solution expressed with the Bessel functions is correct :
y(t) = (a*BesselJ[1/2 , t]+b*BesselJ[-1/2 , t] ) / Sqrt(t)
But the Bessel functions of fractionnal order 1/2 and -1/2 are particular, since they reduces to simpler functions :
BesselJ[1/2 , t] = Sqrt(2/pi)*sin(t)/Sqrt(t)
BesselJ[-1/2 , t] = Sqrt(2/pi)*cos(t)/Sqrt(t)
Then, let A=a*Sqrt(2/pi) and B=b*Sqrt(2/pi) and we obtain :
y(t) = (A*sin(t)+B*cos(t) ) / t
Condition y(pi)=0 implies B=0
Condition y(0)=1 implies A=1 and B=0 leads to the result :
y(t) = sin(t)/t
which is consistent with the solution obtained with the Laplace Transform method.