Question

102–106. Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by F(s) = ∫[0 to ∞] e^(-st) f(t) dt, where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^(-t), the following improper integral is evaluated using integration by parts:
F(s) = ∫[0 to ∞] e^(-st) e^(-t) dt = ∫[0 to ∞] e^(-(s+1)t) dt = 1/(s+1).
Verify the following Laplace transforms, where a is a real number.
104. f(t) = t → F(s) = 1/s²

Accepted Answer

Start with the definition of the Laplace transform for the function $$f(t) = t$$:

$$F(s) = \int_0^{\infty} e^{-st} t \, dt$$ Recognize that this integral involves the product of $$t$$ and an exponential function, which suggests using integration by parts. Recall the integration by parts formula:

$$\int u \, dv = uv - \int v \, du$$ Choose $$u = t so $$that $$du = dt$$, and choose $$dv = e$$^{-st}$$ dt so $$that $$v = \int e$$^{-st}$$ dt = -\frac{1}{s} e$$^{-st}$$. Substitute these into the integration by parts formula: Apply the limits of integration from 0 to $$\infty$$ to the term uv = t$$ \cdot \left(-\frac{1}{s} $$e^{-st}$$ight)$$ and evaluate the remaining integral $$\$$int_0$$^{\infty} \frac{1}{s} $$e^{-st} dt$. Simplify the expression carefully, noting that the exponential term e$$^{-st}$$ tends to zero as t$$ 	o \infty$$ for s$$ \u003e 0$$, and evaluate the definite integrals to express F(s$$)$$ in terms of s$. This will lead to the result F(s$$) = \frac{1}{$$s^2$$}$$.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Definition of the Laplace Transform

Improper Integrals and Convergence

Integration Techniques: Integration by Parts