The discussion focuses on the application of the Laplace transform to solve ordinary differential equations (ODEs) with variable coefficients, specifically the equation 4x y" + 2 y' + y = exp(-x). The Laplace transform of the term xy"(x) is defined and involves integration by parts, leading to an algebraic equation. However, the complexity increases with variable coefficients, making the process less straightforward compared to linear equations with constant coefficients. Participants also address potential typographical errors in the integration process and clarify the interpretation of terms in the context of Laplace transforms.
PREREQUISITESMathematicians, engineering students, and anyone involved in solving differential equations, particularly those interested in advanced techniques for handling variable coefficients using Laplace transforms.
HallsofIvy said: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 /> 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.