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Sahithyan's S3
Sahithyan's S3 — Differential Equations

Laplace Transform

The process of converting a function of real variable tt into a complex variable ss.

F(s)=L{f(t)}=0estf(t)dtF(s) = \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \,\text{d}t

Important formulas

For n0n\ge 0.

L{tn}=n!sn+1\mathcal{L} \{ t^n \} = \frac{n!}{s^{n+1}} L{eat}=1sa\mathcal{L} \{ e^{at} \} = \frac{1}{s-a} L{sin(at)}=as2+a2\mathcal{L} \{ \sin (at) \} = \frac{a}{s^2 + a^2} L{cos(at)}=ss2+a2\mathcal{L} \{ \cos (at) \} = \frac{s}{s^2 + a^2} L{sinh(at)}=as2a2\mathcal{L} \{ \sinh (at) \} = \frac{a}{s^2 - a^2} L{cosh(at)}=ss2a2\mathcal{L} \{ \cosh (at) \} = \frac{s}{s^2 - a^2}

Properties

Linearity

L{cf(t)+g(t)}=cL{f(t)}+L{g(t)}\mathcal{L} \{ cf(t) + g(t) \} = c\mathcal{L}\{ f(t) \} + \mathcal{L} \{ g(t) \}

Shiting

Suppose F(s)F(s) is the Laplace transform of ff, where s>αs \gt \alpha.

L{eatf(t)}=F(sa)\mathcal{L} \{ e^{at} f(t) \} = F(s-a)

For s>α+as \gt \alpha + a.

Time scaling

L{f(αt)}=1αF(sα)\mathcal{L} \{ f(\alpha t) \} = \frac{1}{\alpha} F\left(\frac{s}{\alpha}\right)

Transform with t

L{tnf(t)}=(1)ndndsnF(s)\mathcal{L} \{ t^n f(t) \} = (-1)^n \frac{\text{d}^n}{\text{d}s^n} {F(s)} L{1tf(t)}=sF(x)dx\mathcal{L} \{ \frac{1}{t} f(t) \} = \int_s^\infty F(x)\,\text{d}x

Time shifting

L{u(ta)f(ta)}=easF(s)\mathcal{L}\{ u(t-a)f(t - a) \} = e^{-as}F(s)

Here uu is the unit step function.

Derivatives and Integrals

Suppose F(s)F(s) is the Laplace transform of f(t)f(t).

Derivative

L{f(t)}=sL{f(t)}f(0)\mathcal{L} \{ f'(t) \} = s\mathcal{L} \{ f(t) \} - f(0) L{f(t)}=s2L{f(t)}sf(0)f(0)\mathcal{L} \{ f''(t) \} = s^2\mathcal{L} \{ f(t) \} - sf(0) - f'(0) L{f(n)(t)}=snL{f(t)}k=0n1skfn1k(0)\mathcal{L} \{ f^{(n)}(t) \} = s^n\mathcal{L} \{ f(t) \} - \sum_{k=0}^{n-1} s^k f^{n - 1 - k}(0)

Integral

L{0tf(x)dx}=1sF(s)\mathcal{L} \left\{ \int_0^t f(x) \,\text{d}x \right\} = \frac{1}{s} F(s)

Provided that s0s \neq 0.

Inverse Laplace transform

Reverse of Laplace transform.

L{f(t)}=F(s)    L1{F(s)}=f(t)\mathcal{L}\{ f(t) \} = F(s) \implies \mathcal{L}^{-1} \{ F(s) \} = f(t)

For periodic functions

Suppose f(t)f(t) is a periodic function with period TT. If the laplace transform of f(t)f(t) exists then:

L{f(t)}=11esT0Testf(t)dt\mathcal{L}\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_0^T e^{-st} f(t) \,\text{d}t