Laplace transform for solving ODE with variable coefficients

[angelas]

Can we use laplace transform to solve an ODE with variable coefficients?

Like this one:

4x y" + 2 y' + y = exp (-x)

 

[HallsofIvy]

Theoretically, yes, if you can find the Laplace transform of f(x)y"(x)!

The Laplace transform of xy"(x) is, by definition,
$$\int_0^\infty e^{-st}ty"(t)dt$$
Integrate by parts with $u= e^{-st}t$ so that $du= (e^{-st}- st e^{-st})dt$, dv= y"(t)dt so that v= y'(t). Then
$$-\int_0^\infty (e^{-st}- ste^{-st})y'(t)dt$$
Do that by integration by parts with $u= (e^{-st}- ste^{-st}$ so that $du= -2e^{-st}+ s^2te^{-st}$, dv= y'(t)dt so that v= y. Then
[tex]2y(0)+ \int_0^\infty (2e^{-st}- s^2te^{-st})y(t)dt[/itex]
Write that last as a Laplace transform of y and reduce to an algebraic equation as usual.

Unfortunately, that is not always easy to do with general variable coefficients. The Laplace transform is basically a method for very mechanically solving linear equations with constant coefficients.

 

[angelas]

Theoretically, yes, if you can find the Laplace transform of f(x)y"(x)!

The Laplace transform of xy"(x) is, by definition,
$$\int_0^\infty e^{-st}ty"(t)dt$$
Integrate by parts with $u= e^{-st}t$ so that $du= (e^{-st}- st e^{-st})dt$, dv= y"(t)dt so that v= y'(t). Then
$$-\int_0^\infty (e^{-st}- ste^{-st})y'(t)dt$$
Do that by integration by parts with $u= (e^{-st}- ste^{-st}$ so that $du= -2e^{-st}+ s^2te^{-st}$, dv= y'(t)dt so that v= y. Then
[tex]2y(0)+ \int_0^\infty (2e^{-st}- s^2te^{-st})y(t)dt[/itex]
Write that last as a Laplace transform of y and reduce to an algebraic equation as usual.
Thanks so much.
I think there is a typo here: du = -2e^{-st}+ s^2te^{-st}

it should be du = -2 s e^{-st}+ s^2te^{-st}

and also about the last part why we get 2 y(0). I think it should be zero.

and also how can I write s^2 t e ^(-st) y as the laplace transform of y? is it equal to the laplace transform of s^2ty?

Sorry if I ask too many questions.