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
2y(0)+ \int_0^\infty (2e^{-st}- s^2te^{-st})y(t)dt[/itex]<br /> Write that last as a Laplace transform of y and reduce to an algebraic equation as usual.<br /> <br /> 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&quot;(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&#039;(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
2y(0)+ \int_0^\infty (2e^{-st}- s^2te^{-st})y(t)dt[/itex]<br /> Write that last as a Laplace transform of y and reduce to an algebraic equation as usual.<br /> Thanks so much.<br /> I think there is a typo here: du = -2e^{-st}+ s^2te^{-st}<br /> <br /> it should be du = -2 s e^{-st}+ s^2te^{-st}<br /> <br /> and also about the last part why we get 2 y(0). I think it should be zero.<br /> <br /> 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?<br /> <br /> Sorry if I ask too many questions.