DE: LaPlace

In chapter 6 we will use the LaPlace transform to turn a linear DE into an algebra problem using linearity and a table of LaPlace transforms.

Once we simplify the RHS, we have solved our DE for by just using algebra and a table of formulas. Unfortunately simplifying the RHS in step 5 will take the most work, involving a bunch of algebra plus using the LaPlace transform table. The algebra needed for taking the inverse LaPlace transform will be reviewed in the student created videos below. This algebra for taking the inverse LaPlace transform is also summarized on page 2 of the LaPlace transform table. If you wish to see the rest of this problem, the full solution is available by clicking here . But you can also wait until we cover this in class. But to prepare for this, please see the discussion below and the following:

Take a look at the LaPlace transform table. The table we will use is just 1 page, but there are longer versions on the web (but we will just need the formulas in this 1 page table). To keep track of where we are in the problem, we will use the variable $t$ before we take the LaPlace transform and the variable $s$ after we take the LaPlace transform.

This technique applies to any linear differential equation of any order, but for simplicity, we will start with 2nd order: $y'' + 3y' + 4y = g(t)$

Thus when we take the LaPlace transform of both sides, we will need to be able to calculate ${\cal L}(g(t))$. To do this, we look at column 1 in the LaPlace transform table. For example if $g(t) = e^{at}$, then ${\cal L}(g(t)) ={\cal L}(e^{at}) = \frac{1}{s-a}$. For $a = 2, ~~~{\cal L}(e^{2t}) = \frac{1}{s-2}$.

Note per above we used the variable $t$ before we take the LaPlace transform and the variable after we take the LaPlace transform.

We will first focus on taking the LaPlace transform before we look at the more involved algebra when taking the inverse LaPlace transform.

We will first do a few examples using linearity and the LaPlace transform table to calculate the LaPlace transform of several functions. To use the LaPlace transform table to calculate ${\cal L}(g(t))$, we look at the first column. Note that after taking the LaPlace transform, we get a function of $s$.

Use linearity to calculate the LaPlace transform of $f\left( t \right) = 6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9$.

$LaTeX: {\cal L}(6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9(1)) =6 {\cal L}({{\bf{e}}^{ - 5t}}) + {\cal L}({{\bf{e}}^{3t})} + 5{\cal L}({t^3}) - 9{\cal L}(1)$

$\begin{align*}F\left( s \right) & = 6\frac{1}{{s - \left( { - 5} \right)}} + \frac{1}{{s - 3}} + 5\frac{{3!}}{{{s^{3 + 1}}}} - 9\frac{1}{s}\\ & = \frac{6}{{s + 5}} + \frac{1}{{s - 3}} + \frac{{30}}{{{s^4}}} - \frac{9}{s}\end{align*}$

Use linearity to calculate the LaPlace transform of g(t)=4cos(4t)−9sin(4t)+2cos(10t)

\begin{align*}G\left( s \right) & = 4\frac{s}{{{s^2} + {{\left( 4 \right)}^2}}} - 9\frac{4}{{{s^2} + {{\left( 4 \right)}^2}}} + 2\frac{s}{{{s^2} + {{\left( {10} \right)}^2}}}\\ & = \frac{{4s}}{{{s^2} + 16}} - \frac{{36}}{{{s^2} + 16}} + \frac{{2s}}{{{s^2} + 100}}\end{align*}

Before looking at inverse transform examples, we will first review some algebra.

$$s^2 + 3s + \underline{\hskip 0.2in} - \underline{\hskip 0.2in} + 4 = (s + \underline{\hskip0.2in})^2- \underline{\hskip 0.2in} + 4$$ Hence $s^2 + 3s + 4 = \left(s + {3 \over 2}\right)^2 + {7 \over 4}$

Note to complete the square, you add "0" (in this example, "0" takes the form $\frac{9}{4} - \frac{9}{4}$). When you do algebra, you need to preserve equality. Thus you can add "0".

Hint: Factor out the 5 and be very careful to preserve equality as you modify the above polynomial in order to complete the square.

$$\begin{align*}5s^2 + 20s + 15 & = 5[s^2 + {\color{red} 4} s + 3] = 5[(s + \underline{\hskip0.2in})^2- \underline{\hskip 0.2in} + 3] \\ & = 5[(s + \underline{~{\color{orange} 2}~})^2- \underline{\hskip 0.2in} + 3] \\ & = 5[s^2 + {\color{red} 4}s + \underline{~4~} - \underline{~4~} +3]\\ & = 5[(s + {\color{orange} 2})^2 - 4 + 3] = 5[(s + 2)^2 - 1]\\& = 5(s + 2)^2 - 5 \end{align*}$$

Note that the number ${\color{orange} 2}$ in orange is half the value of the number ${\color{red} 4}$ in red.

NOTE: You must be able to do partial fractions. For a review of partial fractions, please see Partial fractions examples.

We will now focus on calculating the inverse LaPlace transform (which means going from column 2 with LaTeX: F(s)

to column 1 to get $LaTeX: f(t) = {\cal L}^{-1}(F(s))$ .

We will also use the fact that the inverse LaPlace transform is linear plus lots of algebra in order to make your LaTeX: F(s)

look like (a linear combination of) something(s) in column 2.

See page 2 of table of LaPlace transforms to see a summary of the main algebraic techniques. Note we usually focus on the denomator first to determine which formula(s) to use.

Calculate the inverse LaPlace transform of $LaTeX: F(s) = \frac{1}{s^4}$ per below:

To calculate the inverse LaPlace transform of $LaTeX: F(s) = \frac{1}{s^4}$ , you look at the second column and look for the formula that most closely matches it. In this case that would be formula 3:

Comparing the denominators of $LaTeX: F(s) = \frac{1}{s^4}$ and $LaTeX: \frac{n!}{s^{n+1}}$ , we can see that LaTeX: n + 1 = 4

and thus

Letting

in formula 3: $LaTeX: {\cal L}^{-1}\left(\frac{3!}{s^{3+1}}\right) = {\cal L}^{-1}\left(\frac{6}{s^{4}}\right) = t^3$

We have $LaTeX: F(s) = \frac{1}{s^4}$ , but to use the formula, we need $LaTeX: \frac{6}{s^{4}}$ .

Fortunately, we are all powerful when it comes to math. If we want something, we can make it happen, often by multiplying by 1 or adding 0 (so that we don't change the problem). In this case, we will multiply by $\frac{6}{6} = 1$.

Thus $LaTeX: {\cal L}^{-1}\left(\frac{1}{s^4} \right)={\cal L}^{-1}\left(\frac{1}{6}\left(\frac{6}{s^4}\right) \right)=\frac{1}{6}{\cal L}^{-1}\left(\frac{6}{s^{4}}\right)=\frac{1}{6}t^3$

It might sometime feel like you are not all powerful when it comes to math. However, you really are. Even if you make lots of algebraic typos (which your professors also do), you are still better than a computer. It takes a lot of work to program a computer to detect patterns and even then computers are not as good as us in most cases (even IF we make more errors).

Video 16: Ch 6 Inverse LaPlace Transform 7

Calculate the inverse LaPlace transform of $LaTeX: \displaystyle F\left( s \right) = \frac{{1 - 3s}}{{{s^2} + 8s + 21}}$

When calculating the inverse LaPlace transform, look at the denominator. If it factors over the reals, you will need to use partial fractions. But in this example, the quadratic polynomial has complex roots and thus does not factor over the reals. Thus to make it look like the formulas in column 2, you will need to complete the square.

In your video, include the linearity step. Also show the formulas that you use. This example is from Pauls' online notes Section 4-3 : Inverse Laplace Transforms. Click on "Show Answers" for helpful information.

Calculate the inverse LaPlace transform of $LaTeX: \displaystyle H\left( s \right) = \frac{{s + 7}}{{{s^2} - 3s - 10}}$

The denominator in this example factors over the reals. Thus to make it look like the formulas in column 2, you will need use partial fractions.

Sidenote: FYI if you are given initial values for a nonzero LaTeX: t_0

, you can always translate the problem to one where LaTeX: t_0 = 0

and translate back. But all our IVP in ch 6 will have LaTeX: t_0 = 0

We first need the formula for $LaTeX: {\cal L}(y^{(n)})$ where $LaTeX: y^{(n)}$ is the nth derivative of

. This is formula 18, though this formula uses LaTeX: f

instead of LaTeX: y

Formula 18: $LaTeX: {\cal L}(f^{(n)}(t)) = s^nF(s) - s^{n-1}f(0) - ... - f^{(n-1)} (0)$

Since

, we have that $LaTeX: f^{(n)}(t) = y^{(n)}$ and the initial values are $LaTeX: y(0) = f(0), y'(0) = f'(0), ..., y^{(n-1)} (0) = f^{(n-1)}(0)$ .

Formula 18': $LaTeX: {\cal L}(y^{(n)}) = s^n{\cal L}(y) - s^{n-1}y(0) - ... - y^{(n-1)} (0)$

Note that if we take the LaPlace transform of the $LaTeX: y^{(n)}$ = nth derivative of

, we get a degree LaTeX: n

polynomial.

$LaTeX: {\cal L}(y') = s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y'') = s^2( \underline{\hskip 1cm}) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y''') = s^3( \underline{\hskip 1cm}) -s^2( \underline{\hskip 1cm}) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y'''') = s^4( \underline{\hskip 1cm}) - s^3( \underline{\hskip 1cm}) -s^2( \underline{\hskip 1cm}) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}$

Note the pattern. Lets now fill in the first blank for each of these. Note that the coefficient of LaTeX: s^n

in Formula 18 is $LaTeX: {\cal L}(y)$ .

$LaTeX: {\cal L}(y') = s {\cal L}(y) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y'') = s^2 {\cal L}(y) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y''') = s^3 {\cal L}(y) -s^2( \underline{\hskip 1cm}) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}\\~ \\ {\cal L}(y'''') = s^4 {\cal L}(y) - s^3( \underline{\hskip 1cm}) -s^2( \underline{\hskip 1cm}) -s ( \underline{\hskip 1cm}) -\underline{\hskip 1cm}$

The remaining blanks are filled in with initial values, starting with y(0). As the degree of s decreases, the derivative of y (evaluated at 0) increases. Note we always end with subtracting a constant term as we run out of initial values at the same time we run out of LaTeX: s

's.

$LaTeX: {\cal L}(y') = s{\cal L}(y) - y(0)\\~\\ {\cal L}(y'') = s^2{\cal L}(y) - sy(0) - y'(0)\\~\\ {\cal L}(y''') = s^3{\cal L}(y) - s^2y(0) - sy'(0) - y''(0)\\~\\ {\cal L}(y'''') = s^4{\cal L}(y) - s^3y(0) - s^2y'(0) - sy''(0) - y'''(0)$

Note this is the example in the first post. We wish to solve for LaTeX: y

. It is fairly quick to solve for LaTeX: y

in terms of an inverse LaPlace transform, but calculating the inverse LaPlace transform will take a few steps.

In this video solve for LaTeX: y

in terms of an inverse LaPlace transform, but do not calculate the inverse LaPlace transform. We will save that part of later videos.

Note that 0 does not appear in the LaPlace transform table, but for any linear function $LaTeX: {\cal L}(0) = {\cal L}(0 \cdot 0) = 0{\cal L}(0) = 0$ . Thus we get 0 on the RHS after taking the LaPlace transform of 0.

$LaTeX: [s^2{\cal L}(y) - sy(0) - y'(0)] + 3[s{\cal L}(y) - y(0)] + 4{\cal L}(y) = 0$

$LaTeX: s^2{\cal L}(y) - 5s - 6 + 3s{\cal L}(y) - 15 + 4{\cal L}(y) = 0\\ ~\\ [s^2 + 3s + 4]{\cal L}(y) = 5s + 21\\ ~\\ {\cal L}(y) = {5s + 21 \over s^2 + 3s + 4}$

$LaTeX: y = {\cal L}^{-1}({\cal L}(y)) = {\cal L}^{-1}\left( {5s + 21 \over s^2 + 3s + 4}\right)$

We have now solved the differential equation for LaTeX: y

, but we now want to simplify the RHS by calculating the inverse LaPlace transform. That will be a lot of work. Thus we will save that for the next videos.

Part 2: Calculate $LaTeX: y = {\cal L}^{-1}({\cal L}(y)) = {\cal L}^{-1}\left( {5s + 21 \over s^2 + 3s + 4}\right)$

In part 1, we solved the following IVP for LaTeX: y

in terms of an inverse LaPlace transform.

$LaTeX: y = {\cal L}^{-1}({\cal L}(y)) = {\cal L}^{-1}\left( {5s + 21 \over s^2 + 3s + 4}\right)$

we now want to simplify the RHS by calculating the inverse LaPlace transform. Per the previous videos, we first see if the denominator factors over the reals. It does not, so we need to complete the square. Lots more algebra will then be needed to make it look like a linear combination of 2 formulas in the LaPlace transform table. Note this algebra involves adding a form of 0 and multiplying by a form of 1. To see the solution worked out, click here

Video 22: Ch 6 Solve $LaTeX: y'' - 10y' + 9y = 5t,\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,\,y'\left( 0 \right) = 2$

Part 1 In this video solve for LaTeX: y

in terms of an inverse LaPlace transform, but do not calculate the inverse LaPlace transform. We will save that part for the next video.

Some notation: Many books use $LaTeX: Y(s) = {\cal L}(y)$ . You are welcome to use either. Since we are solving for

, I like $LaTeX: {\cal L}(y)$ so that I can see the variable I am solving for. LaTeX: Y(s)

is nice to remind one that we have taken the LaPlace transform and thus $LaTeX: {\cal L}(y) = Y(s)$ is a function of LaTeX: s

. Recall that to remind us which column we should use, we use the variable LaTeX: t

for before we take the LaPlace transform and the variable LaTeX: s

for after we have taken the LaPlace transform.

Video 23: Ch 6 Solve $LaTeX: y'' - 10y' + 9y = 5t,\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,\,y'\left( 0 \right) = 2$

$LaTeX: y'' - 10y' + 9y = 5t,\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,\,y'\left( 0 \right) = 2$

$LaTeX: y = {\cal L}^{-1}{\cal L}(y) ={\cal L}^{-1}(Y\left( s \right)) = {\cal L}^{-1}\left(\frac{5}{{{s^2}\left( {s - 9} \right)\left( {s - 1} \right)}} + \frac{{12 - s}}{{\left( {s - 9} \right)\left( {s - 1} \right)}}\right)$

we now want to simplify the RHS by calculating the inverse LaPlace transform. Note the first step is to add the two terms (after getting common denominator) as it will be less work to calculate the inverse LaPlace transform of the sum rather than doing 2 separate inverse LaPlace transforms. Whether one should combine terms or not depends on the problem, but combining is often less work.