Laplace Transform
We will cover following topics
Laplace And Inverse Laplace Transforms
Let a function \(f\) be defined for \(t \geq 0\). Then, Laplace Transform of \(f\) is defined as:
\[F(s)=\mathcal{L}(f(t))=\int_{0}^{\infty} e^{-s t} f(t) d t\]for those values of \(s\) where this integral exists.
Also, the inverse Laplace Transform is defined as:
\[f(t)=\mathcal{L}^{-1}(F(s))\]First Shifting Theorem: If \(\mathcal{L}(f(t))=F(s)\), then \(\mathcal{L}\left(e^{a t} f(t)\right)=F(s-a)\).
Laplace Transforms Of Elementary Functions
The unit step function \(u_{a}(t)\) is defined as:
\[u_{a}(t)=\left\{\begin{array}{ll}{0,} & {t< a} \\ {1,} & {t>a}\end{array}\right.\]Laplace Transform of \(u_{a}(t)\),
\[\mathcal{L}\left(u_{a}(t)\right)=e^{-a s} / s\]Second Shifting Theorem: If \(\mathcal{L}(f(t))=F(s)\), then
\[\mathcal{L}\left(u_{a}(t) f(t-a)\right)=e^{-a s} F(s)\]Theorem: Let \(f(t)\) be continuous for \(t \geq 0\) and is of exponential order. Further, suppose that \(f\) is differentiable with \(f'\) piecewise continuous in \([0, \infty)\). Then,
\[\mathcal{L}\left(f^{\prime}\right)=s \mathcal{L}(f)-f(0)\]Also,
\(\mathcal{L}\left(f^{(n)}\right)\)=\(s^{n} \mathcal{L}(f)\)-\(s^{n-1} f(0)\)-\(s^{n-2} f^{(1)}(0)\)-\(\cdots\)-\(f^{(n-1)}(0)\)
Theorem: Let \(F(s)\) be the Laplace Transform of \(f\). If \(f\) is piecewise continuous in \([0, \infty)\) and is of exponential order, then
\[\mathcal{L}\left(\int_{0}^{t} f(\tau) d \tau\right)=\dfrac{F(s)}{s}\]Theorem: Let \(F(s)\) be the Laplace Transform of \(f\), then
\[\mathcal{L}(-t f(t))=F^{\prime}(s)\]Theorem: Let \(F(s)\) be the Laplace Transform of \(f\) and the limit of \(f(t)/t\) exists as \(t \rightarrow 0^{+}\), then
\[\mathcal{L}\left(\dfrac{f(t)}{t}\right)=\int_{s}^{\infty} F(p) d p\]Initial Value Problems
Theorem for Periodic Functions: Let \(f :[0, \infty) \rightarrow \mathbb{R}\) be a periodic function with period \(T>0\), i.e., \(f(t+T)=f(t) \text{ } \forall \text{ } t \geq 0\). Then,
\[F(s)=\dfrac{\int_{0}^{T} f(t) e^{-s t} d t}{1-e^{-s T}}\]Convolution: Let \(f\) and \(g\) be two functions defined in \([0, \infty)\).
Then, the convolution of \(f\) and \(g\),is defined by:
\[(f * g)(t)=\int_{0}^{\infty} f(\tau) g(t-\tau) d \tau\]Convolution Theorem: The convolution \(f*g\) has the Laplce Transform property:
\[\mathcal{L}((f * g)(t))=F(s) G(s)\]PYQs
Laplace And Inverse Laplace Transforms
1) Find the Laplace transforms of \(t^{-1/2}\) and \(t^{-1/2}\). Prove that the Laplace transform of \(t^{n+1/2}\), where \(n\in N\), is
\[\dfrac{\Gamma{(n+1+\dfrac{1}{2})}}{s^{n+1+\dfrac{1}{2}}}\][2019, 10M]
2) Find the Laplace transform of \(f(t)=\dfrac{1}{\sqrt{t}}\).
[2018, 5M]
3) Find the inverse Laplace transform of \(\dfrac{5s^2+3s-16}{(s-1)(s-2)(s-3)}\).
[2018, 5M]
4) Obtain Laplace Inverse transform of \(\left\{\ln \left(1+\dfrac{1}{s^{2}}\right)+\dfrac{s}{s^{2}+25} e^{-5 s}\right\}\).
[2015, 6M]
5) Find the inverse Laplace transform of \(F(s)=1 n\left(\dfrac{s+1}{s+s}\right)\).
[2009, 20M]
Initial Value Problems
1) Solve the initial value problem
\(y''-5y'+4y=e^{2t}\)
\(y(0)=\dfrac{19}{20}, y'(0)=\dfrac{8}{3}\)
[2018, 13M]
2) Solve the following initial value problem using Laplace transform:
\[\dfrac{d^{2} y}{d x^{2}}+9 y=r(x), y(0)=0, y^{\prime}(0)=4\]where
\[r(x)=\left\{\begin{array}{ll}{8 \sin x} & {\text { if } 0< x < \pi} \\ {0} & {\text { if } x \geq \pi}\end{array}\right.\][2017, 17M]
3) Using Laplace transformation, solve the following: \(y^{\prime \prime}-2 y^{\prime}-8 y=0\), \(y(0)=3\), \(y^{\prime}(0)=6\).
[2016, 10M]
4) Using Laplace transform, solve \(y^{\prime \prime}+y=t\), \(y(0)=1\), \(y^{\prime}(0)=-2\).
[2015, 6M]
5) Solve the initial value problem \(\dfrac{d^{2} y}{d t^{2}}+y=8 e^{-2 t} \sin t\), \(y(0)=0\), \(y^{\prime}(0)=0\) by using Laplace transform.
[2014, 20M]
6) By using Laplace transform method, solve the differential equation \(\left(D^{2}+n^{2}\right) x=a \sin (n t+\alpha)\) \(\left( D^{2}=\dfrac{d^{2}}{d t^{2}} \right)\) subject to the initial conditions \(x=0\) and \(\dfrac{d x}{d t}=0\), in which \(a\), \(n\) and \(\alpha\) are constants.
[2013, 15M]
7) Using Laplace transforms, solve the initial value problem \(y^{\prime \prime}+2 y^{\prime}+y=e^{-t}\), \(y(0)=-1\), \(y^{\prime}(0)=1\).
[2012, 12M]
8) Use Laplace transform method to solve the following initial value problem: \(\dfrac{d^{2} x}{d t^{2}}-2 \dfrac{d x}{d t}+x=e^{t}\), \(x(0)=2\) and \(\left.\dfrac{d y}{d t}\right \vert_{t=0}=-1\).
[2011, 15M]
9) Using Laplace transform, solve the initial value problem \(y^{\prime \prime}-3 y^{\prime}+2 y=4 t+e^{3 t}\), \(y(0)=1\), \(y^{\prime}(0)=-1\).
[2008, 15M]