Find solution with initial value

[invisible_man]

y'' - ty' + y = 1, y(0) = 1, y'(0) =2

I'm trying to solve by Laplace Transform and I got stuck at the end.Anyone can help me to solve it?

 

[matematikawan]

Did you compute the Laplace transform of ty'(t) correctly? I think the Laplace transform of this term and computing the Laplace inversion later are the only diffuculty that I foresee.

 

[invisible_man]

Did you compute the Laplace transform of ty'(t) correctly? I think the Laplace transform of this term and computing the Laplace inversion later are the only diffuculty that I foresee.

The Laplace Transform of ty' = -L[y]'

 

[matematikawan]

I think so. Except that the RHS differentiation is wrt s, i.e. -\frac{dF}{ds}= -\frac{d}{ds}(sY-1)

 

[kosovtsov]

Your ODE is not an ODE with constant coefficients, so it can not be solved by Laplace Transform.

An integrating factor to your ODE

\mu=t\exp(-\frac{t^2}{2})

and corresponding first integral is

I=\exp(-\frac{t^2}{2})(t\frac {d y}{d t}-y+1)+C1

in sense that

\frac {d I}{d t}=t\exp(-\frac{t^2}{2})(\frac {d^2 y}{d t^2}-t\frac {d y}{d t}+y-1),

what is equivalent to your ODE.


Substituting t=0 in I we get

-y(0)+1=C1

and as y(0)=1 then

C1=0.

Now you have to solve first order ODE (I=0)

t\frac {d y}{d t}-y(t)+1=0.

Its solution is

y(t) = tC2+1

Your initial conditions lead to the following solution

y(t) = 2t+1

 

[matematikawan]

Your ODE is not an ODE with constant coefficients, so it can not be solved by Laplace Transform.

Some linear DE with variable coefficients can be solve via Laplace transform. It all depends on whether we can solve for Y(s) and later invert it. In this particular case, the transformed equation will be a linear first order DE in Y(s) which I think can be solve for Y(s) quite easily. But whether it can be invert easily, need to look at the expression first.

 

[kosovtsov]

Some linear DE with variable coefficients can be solve via Laplace transform

Can you give an example, please?

 

[matematikawan]

I remember some time ago somebody try to solve Bessel equation via Laplace transform.
https://www.physicsforums.com/showthread.php?t=389727"
But that person didn't proceed to see them through.

But for the Bessel equation of order zero, it can be done.
x²y''+xy'+x²y = 0
xy''+y'+xy = 0

Apply Laplace transform and solve for Y(s)
-\frac{d}{ds}(s^2Y-as-b) + sY-a-\frac{dY}{ds}

\frac{dY}{ds}=-\frac{sY}{s^2+1}

Y(s)=\frac{A}{\sqrt{s^2+1}}

The solution will be the inversion of Y(s).