// Numbas version: exam_results_page_options {"name": "Laplace Transforms", "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], []], "questions": [{"name": "Laplace of constants and powers of t", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d"], "tags": ["rebelmaths"], "advice": "

See 'show steps'.

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Find $L\\{\\var{a} \\}$.

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$L\\{k\\}=\\frac{k}{s}$

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Find $L\\{\\var{b}t \\}$.

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$L\\{t\\}=\\frac{1}{s^2}$

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Find $L\\{\\var{a}+\\var{b}t \\}$.

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$L\\{k\\}=\\frac{k}{s}$

$L\\{t\\}=\\frac{1}{s^2}$

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Find $L\\{t^\\var{a}\\}$.

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$L\\{t^n\\}=\\frac{n!}{s^{n+1}}$

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Find $L\\{\\var{a}t^\\var{c}+\\var{b}t^\\var{d}\\}$.

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$L\\{t^n\\}=\\frac{n!}{s^{n+1}}$

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You may use a table of Laplace transforms in order to answer the following questions.

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Laplace of constants and powers of t

rebelmaths

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Ugur's copy of Laplace transform #3 - powers of t by exp", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Marie Nicholson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1799/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

What is the Laplace transform for the function:

\$x(t)=\\var{a}t^{\\var{n}}e^{-\\var{k}t}+\\var{b}.\$

", "advice": "

By linearity of Laplace Transform we have

\\[X(s) = \\mathcal L\\left\\{\\var{a}t^{\\var{n}}e^{-\\var{k}t}+\\var{b}\\right\\} = \\var{a} \\mathcal L\\left\\{t^{\\var{n}}e^{-\\var{k}t}\\right\\}+\\mathcal L\\left\\{\\var{b}\\right\\}.\\]

Then apply First Shift Theorem to $\\mathcal L\\left\\{t^{\\var{n}}e^{-\\var{k}t}\\right\\}$, and use trasmform table for $\\mathcal L\\left\\{\\var{b}\\right\\}$ to get the answer.

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\$X(s)=\$ [[0]]

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What is the Laplace transform for the function:

\$x(t)=\\var{a}t^{\\var{n}}+\\var{b}e^{-\\var{k}t}.\$

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\$X(s)=\$ [[0]]

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What is the Laplace transform for the function:

\$x(t)=\\var{a}\\sin({\\var{n}t})+\\var{b}\\cos({\\var{k}t}).\$

", "advice": "

By the linearity of Laplace Transform:

\\[X(s)=\\mathcal L\\{\\var{a}\\sin({\\var{n}t})+\\var{b}\\cos({\\var{k}t})\\} = \\var{a}\\mathcal L\\{\\sin({\\var{n}t})\\}+\\var{b}\\mathcal L\\{\\cos({\\var{k}t})\\}.\\]

From Laplace Transform table one can read

\\[\\mathcal L\\{\\sin({\\var{n}t})\\} = \\frac{\\var{n}}{s^2 + \\var{n}^2} \\quad \\mbox{ and } \\quad \\mathcal L\\{\\cos({\\var{n}t})\\} = \\frac{s}{s^2 + \\var{k}^2}.\\]

Hence

\\[X(s)=\\mathcal L\\{\\var{a}\\sin({\\var{n}t})+\\var{b}\\cos({\\var{k}t})\\} = \\var{a}\\mathcal L\\{\\sin({\\var{n}t})\\}+\\var{b}\\mathcal L\\{\\cos({\\var{k}t})\\} = \\frac{\\simplify{{a}*{n}}}{s^2 + \\simplify{{n}^2}} + \\frac{\\var{b}s}{s^2 + \\simplify{{k}^2}}.\\]

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\$X(s)=\$ [[0]]

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Solve the differential equation

\\[\\frac{d^2q}{dt^2}+\\var{a}\\frac{dq}{dt}+\\var{b}q(t)=\\var{c}e^{-\\var{d}t}\\quad \\mbox{where} \\quad q(0)=\\var{f}\\quad \\mbox{and} \\quad q'(0)=\\var{g}\\]

using Laplace Transform method.

", "advice": "

a) We are given the ODE

\\[\\frac{d^2q}{dt^2}+\\var{a}\\frac{dq}{dt}+\\var{b}q(t)=\\var{c}e^{-\\var{d}t}\\quad \\mbox{where} \\quad q(0)=\\var{f}\\quad \\mbox{and} \\quad q'(0)=\\var{g}.\\]

Using Laplace Transform Table and teh rule for Laplace Transform of derivatives we transfor the ODE into:

\\[s^2Q(s)-sq(0)-q'(0)+\\var{a}(s(Q(s)-q(0))+\\var{b}Q(s)=\\frac{\\var{c}}{s+\\var{d}}.\\]

Pluging in initial values we get :

\\[s^2Q(s)-\\var{f}s-\\var{g}+\\var{a}sQ(s)-\\simplify{{a}*{f}}+\\var{b}Q(s)=\\frac{\\var{c}}{s+\\var{d}}.\\]

Re-arrange to get:

\\[s^2Q(s)+\\var{a}sQ(s)+\\var{b}Q(s)=\\frac{\\var{c}}{s+\\var{d}}+\\var{f}s+\\simplify{{g}+{a}*{f}}.\\]

Take out $Q(s)$ as a common factor on the left-hand side:

\\[(s^2+\\var{a}s+\\var{b})Q(s)=\\frac{\\var{c}+(\\var{f}s+\\simplify{{g}+{a}*{f}})(s+\\var{d})}{s+\\var{d}}.\\]

Make $Q(s)$ the subject:

\\[Q(s)=\\frac{\\simplify{{f}s^2+({a}*{f}+{g}+{d}*{f})s+(({g}+{f}*{a})*{d}+{c})}}{(s+\\var{d})(s^2+\\var{a}s+\\var{b})}.\\]

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Find the Laplace transform of the following differential equation and express it \$Q(s)\$ as a single fraction:

\$\\frac{d^2q}{dt^2}+\\var{a}\\frac{dq}{dt}+\\var{b}q(t)=\\var{c}e^{-\\var{d}t}\$ where \$q(0)=\\var{f}\$ and \$q'(0)=\\var{g}\$

\$Q(s)=\$ [[0]]

OPTIONAL (This could be tricky!):

Find the inverse transform of $Q(s)$.

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Determine the inverse Laplace Transform of the following using completion of the square.

\$F(s)=\\dfrac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\$

", "advice": "

We have

\$F(s)=\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\$.

By completing the square (on the denominator) we get:

\$F(s)=\\frac{\\var{B}s+\\var{C}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}\$.

Re-grouping the numerator gives:

\$F(s)=\\frac{\\var{B}(s+\\var{b1})-\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} - \\frac{\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} - \\frac{\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}\\cdot\\frac{\\sqrt{\\simplify{{c1}-{b1}^2}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} \$

Now, the expression above is easy enough to hunt from Laplace Transform table:

\$f(t)=\\var{B}e^{-\\var{b1}t}\\cos\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)+\\frac{-\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}e^{-\\var{b1}t}\\sin\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)\$

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Write down the inverse Laplace transform

\$\\mathscr{L}^{-1}\\{F(s)\\}=\$ [[0]]

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", "advice": "

We start with the differential equation:

\\[\\frac{d^2x}{dt^2}+\\var{a}\\frac{dx}{dt}+\\var{b}x(t)=\\var{c}e^{-\\var{d}t} \\quad \\mbox{ where }\\quad x(0)=\\var{f} \\mbox{ and } x'(0)=\\var{g}.\\]

If take Laplace transform of both sides, and apply linearity we get

\\[\\mathcal L\\left\\{\\frac{d^2x}{dt^2}\\right\\}+\\var{a}\\mathcal L\\left\\{\\frac{dx}{dt}\\right\\}+\\var{b}\\mathcal L\\left\\{x(t)\\right\\}=\\var{c}\\mathcal L\\left\\{e^{-\\var{d}t}\\right\\}.\\]

Then, using Laplace Trasnform table, and applying the Laplace Transform of Derivatives rule we get

\\[s^2X(s)-sx(0)-x'(0)+\\var{a}(s(X(s)-x(0))+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}.\\]

Plug-in the initial value conditions to get:

\\[s^2X(s)-\\var{f}s-\\var{g}+\\var{a}sX(s)-\\simplify{{a}*{f}}+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}.\\]

Then collect $X$ terms on one side to get:

\\[s^2X(s)+\\var{a}sX(s)+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}+\\var{f}s+\\simplify{{g}+{a}*{f}}.\\]

Take into $X$ common paranthesis:

\\[(s^2+\\var{a}s+\\var{b})X(s)=\\frac{\\var{c}+(\\var{f}s+\\simplify{{g}+{a}*{f}})(s+\\var{d})}{s+\\var{d}}.\\]

Finally make $X$ the subject:

\\[X(s)=\\frac{\\simplify{{f}s^2+({a}*{f}+{g}+{d}*{f})s+(({g}+{f}*{a})*{d}+{c})}}{(s+\\var{d})(s^2+\\var{a}s+\\var{b})}.\\]

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Given the differential equation:

\$\\dfrac{d^2x}{dt^2}+\\var{a}\\dfrac{dx}{dt}+\\var{b}x=\\var{c}e^{-\\var{d}t}\$ where \$x(0)=\\var{f}\$ and \$x'(0)=\\var{g}\$

Take the Laplace transform of the equation,
Substitute in the conditions,
Rearrange for \$\\mathscr{L}(x)\$ and then express the right-hand side as a single fraction.
Enter this fraction in the box below.

\$\\mathscr{L}(x)=\$ [[0]]

Find the inverse Laplace transform of the expression you found in the previous part (this can be challenging).

"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Ugur's copy of Laplace transform: irreducible quadratic factor", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Marie Nicholson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1799/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

The Laplace Transfor of a function $x(t)$ is given by

\\[X(s)=
\\frac{\\simplify{{A} + {B}}s^2 + \\simplify{{A}*{b1}*2 +{B}*{a1} + {C}}s + \\simplify{{A}*{c1}+{C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}}.\\]

", "advice": "

a) Observe that

\\[\\frac{\\simplify{{A}+{B}}s^2+\\simplify{{A}*{b1}*2 + {B}*{a1} + {C}}s + \\simplify{{A}*{c1} + {C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}} =\\frac{\\var{A}s^2+\\simplify{{A}*{b1}*2}s + \\simplify{{A}*{c1}} +\\var{B}s^2 + \\simplify{{B}*{a1}}s +\\var{C}s + \\simplify{{C}*{a1}}}{s^3 +\\simplify{{a1}+{{b1}*2}}s^2 +\\simplify{{c1}+{{a1}*{b1}*2}}s + \\simplify{{a1}*{c1}}}= \\\\[6mm]
=\\frac{\\var{A}(s^2+\\simplify{{b1}*2}s+\\var{c1}) +(s+\\var{a1})(\\var{B}s+\\var{C})}{(s+\\var{a1})(s^2+\\simplify{{b1}*2}s+\\var{c1})} =\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\]

b)
We established
\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\]

in the previous part.

By Completing the square we get:

\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}.\\]

Making it further resemble a transform in the Laplace Transfrom table we need to further manipulate the numerator of the second fraction, and then decompose it fruther:

\\[X(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}(s+\\var{b1})-\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} -\\frac{\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}\\cdot\\frac{\\sqrt{\\simplify{{c1}-{b1}^2}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}.\\]

Now, reading from the Fourier Transform table:

\\[x(t)=\\var{A}e^{\\var{a1}t}+\\var{B}e^{-\\var{b1}t}\\cos\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)+\\frac{-\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}e^{-\\var{b1}t}\\sin\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)\\]

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Find the partial fraction decomposition of $X(s)$.

$X(s) = $

[[0]]

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Write down the inverse Laplace transform

\$x(t)=\$ [[0]]

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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Solve the differential equation
\\[x^{\\prime\\prime} = -\\simplify{{k}*{k}}x - \\var{b}x^\\prime \\quad \\mbox{ with the initial conditions } \\quad x(0) = \\var{l} \\, \\mbox{ and }\\, x^\\prime(0) = 0\\]

", "advice": "

By re-arranging the given ODE we get:

\\[x^{\\prime\\prime} +\\var{b}x^\\prime + \\simplify{{k}*{k}}x = 0. \\]

Then, let $X=\\mathcal{L}\\{x\\}$, and apply Laplace Transform to get

\\[(s^2X-sx(0)-x^\\prime(0)) +\\var{b}(sX-x(0)) +\\simplify{{k}*{k}}X=0.\\]

Then plugging in the initial values:

\\[(s^2X-\\var{l}s) +\\var{b}(sX-\\var{l}) +\\simplify{{k}*{k}}X=0.\\]

By re-arranging, we get

\\[X=\\frac{\\var{l}s + \\simplify{2*{k}{l}}}{s^2+\\var{b}s +\\simplify{{k}*{k}}} = \\frac{\\var{l}(s +\\var{k}) + \\simplify{{l}*{k}}}{(s+\\var{k})^2} = \\frac{\\var{l}}{s+\\var{k}} + \\frac{\\simplify{{l}*{k}}}{(s+\\var{k})^2}.\\]

Finally, by the transform table and the property $\\mathcal{L}\\left\\{t^n g(t)\\right\\} = (-1)^n\\frac{d^nG}{ds^n}$ (where $\\mathcal L\\{g(t)\\} = G(s)$ (Note that one can use the First Shift Theorem instead of this property here).

\\[x(t) =\\var{l}e^{-4t} + \\simplify{{l}*{k}}te^{-4t} \\]

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Take the Laplace transform of the given ODE and write down the transformed equation:

[[0]]

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Re-arrange the transfomed equation to isolate $X$.

$X=$[[0]]

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Find the partial fractions decomposition of the rational function you found above.

$X=$[[0]]

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Find the inverse transform of $X(s)$:

$x(t)=$[[0]]

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