Laplace Transforms

Non-homogeneous differential equations can be solved using Laplace transforms. The homogeneous solution isn't found first and the function doesn't have to be continuous, but the problems has to have initial values.

The Essentials

The definition of Laplace transform:

\[ F(s)=\mathscr{L}(f(t))=\int_0^\infty e^{-st}f(t)dt \]

Here are the basic Laplace transforms:

\[ \mathscr{L}(k)=\frac{k}{s} \]
\[ \mathscr{L}(t^n)=\frac{n!}{s^{n+1}} \]
\[ \mathscr{L}(e^{at})=\frac{1}{s-a} \]
\[ \mathscr{L}(e^{i\omega t})=\frac{1}{s-i\omega} \]
\[ \mathscr{L}(\sin(at))=\frac{a}{s^2+a^2} \]
\[ \mathscr{L}(\cos(at))=\frac{s}{s^2+a^2} \]
\[ \mathscr{L}(\sinh(at))=\frac{a}{s^2-a^2} \]
\[ \mathscr{L}(\cosh(at))=\frac{s}{s^2-a^2} \]
\[ \mathscr{L}(\mathscr{U}(t-a))=\frac{1}{s}e^{-as} \]
\[ \mathscr{L}(e^{at}f(t))=F(s-a)\tag{First Shifting Theorem} \]
\[ \mathscr{L}(tf(t))=-\frac{dF}{ds} \]
\[ \mathscr{L}(t^nf(t))=(-1)^n\frac{d^nF}{ds^n} \]
\[ \mathscr{L}\left(\int_{0}^{t}f(x)dx\right)=\frac{F(s)}{s} \]
\[ \mathscr{L}\left(\int_{0}^{t}f(t-\tau)g(\tau)d\tau\right)=F(s)G(s) \]
\[ \mathscr{L}\left(\frac{f(t)}{t}\right)=\int_s^\infty F(\sigma)d\sigma \]

To solve a differential equation, the equation is converted to Laplace space. Once there it is solved for F(s). Then it is usually necessary to simplify fractions, often with partial fraction expansion. Then, the inverse Laplace is calculated which gives the solution. To Take the Laplace transform of a whole equation these formulas are used:

\[ \mathscr{L}(f(t))=F(s) \]
\[ \mathscr{L}(f'(t))=sF(s)-f(0) \]
\[ \mathscr{L}(f''(t))=s^2F(s)-sf(0)-f'(0) \]
\[ \mathscr{L}(f^{(n)}(t))=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\ldots-f^{(n-1)}(s) \]
\noindent Remember that constants can be factored out of Laplace transforms:
\[ \mathscr{L}(cf(t))=c\mathscr{L}(f(t)) \]

Example

Solve this equation using Laplace transforms:

\[ \begin{cases} y''-2y'+y=1\\ y(0)=-1\\ y'(0)=1 \end{cases} \]

First, we will use the formulas for Laplace transforms of derivatives of the function:

\[ \mathscr{L}(y)=Y(s) \]
\[ \mathscr{L}(y')=sY(s)-y(0) \]
\[ \mathscr{L}(y'')=s^2Y(s)-sy(0)-y'(0) \]

We use these formulas to take the Laplace transform of the original equation:

\[ y''-2y'+y=1 \]
\[ s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+Y(s)=\frac{1}{s} \]

Using the initial conditions, we get:

\[ s^2Y(s)+s-1-2sY(s)-2+Y(s)=\frac{1}{s} \]
\[ s^2Y(s)-2sY(s)+Y(s)+s-3=\frac{1}{s} \]

Now, we solve for Y(s):

\[ Y(s)(s^2-2s+1)=\frac{1}{s}-s+3 \]
\[ Y(s)=\frac{1}{s(s^2-2s+1)}+\frac{-s+3}{(s^2-2s+1)} \]

For the first term on the right hand side, we can use partial fraction expansion to re-write the fraction. The second term can be split into two fraction:

\[ Y(s)=\frac{1}{s(s^2-2s+1)}+\frac{-s+3}{(s^2-2s+1)} \]
\[ Y(s)=\frac{1}{s}-\frac{1}{s-1}+\frac{1}{(s-1)^2}+\frac{-s}{(s-1)^2}+\frac{3}{(s-1)^2} \]
\[ Y(s)=\frac{1}{s}-\frac{1}{s-1}+\frac{1-s}{(s-1)^2}+\frac{3}{(s-1)^2} \]
\[ Y(s)=\frac{1}{s}-\frac{1}{s-1}-\frac{1}{s-1}+\frac{3}{(s-1)^2} \]
\[ Y(s)=\frac{1}{s}-\frac{2}{s-1}+\frac{3}{(s-1)^2} \]

Now we can take the inverse Laplace transform to get the solution:

\[ y(t)=1-2e^t+3te^t \]

Practice

Solve these equations using Laplace transforms. If you get stuck, try working backwards from the answers:

  1. \[ \begin{cases} y'''-y'=3e^t\\ y(0)=6\\ y'(0)=\frac{1}{2}\\ y''(0)=6 \end{cases} \]
  2. \[ \begin{cases} y''+y=\frac{1}{2}\cos(t)-2\\ y(0)=0\\ y'(0)=1 \end{cases} \]
  3. \[ \begin{cases} y''+2y'+y=-x^2\\ y(0)=-4\\ y'(0)=5 \end{cases} \]
  4. \[ \begin{cases} f'(t)+f(t)-2\int_0^tf(\tau)d\tau=t\\ f(0)=0 \end{cases} \]

Solutions:

  1. \( y(t)=3+e^t+2e^{-t}+\frac{3}{2}te^t \)
  2. \( y(t)=\sin(t)+2\cos(t)+\frac{1}{2}t\sin(t)-2 \)
  3. \( y(x)=2e^{-x}+3xe^{-x}-x^2+4x-6 \)
  4. \( f(t)=-\frac{1}{2}+\frac{1}{3}e^t+\frac{1}{6}e^{-2t} \)

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