University Differential Equations — The Laplace Transform Worksheet

All 15 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Laplace Transforms from the Definition and from a Table

   The Laplace transform of a function $f$ defined for $t\ge 0$ is $ℒ\{f(t)\}=F(s)={\displaystyle {\int }_0^{\infty }{e}^{-st}f(t)dt}$, for those $s$ at which the improper integral converges. Use this definition to find $F(s)$ for $$f(t)=\{\begin{array}{cc}t,\hfill & 0\le t<1,\hfill \\ 1,\hfill & t\ge 1,\hfill \end{array}$$ and state the real values of $s$ for which it exists.

2. Q2Laplace Transforms from the Definition and from a Table

   Use linearity and the pairs $$ℒ\{1\}=\frac{1}{s},ℒ\{t^n\}=\frac{n!}{s^{n+1}},ℒ\{e^{at}\}=\frac{1}{s-a},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2}$$ to find the Laplace transform of each function.

   1. $f(t)=3t^2-4e^{-t}+2\sin 5t$
   2. $g(t)={(1+e^{2t})}^2$
   3. $h(t)={\cos }^{2}2t$
3. Q3Inverse Transforms and Partial Fractions

   Using the pairs $$ℒ\{t^n\}=\frac{n!}{s^{n+1}},ℒ\{e^{at}\}=\frac{1}{s-a},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2},$$ find each inverse transform.

   1. ${ℒ}^{-1}\{{\displaystyle \frac{4}{s^3}}-{\displaystyle \frac{6}{s-5}}\}$
   2. ${ℒ}^{-1}\left\{{\displaystyle \frac{2s+7}{s^2+9}}\right\}$
4. Q4Inverse Transforms and Partial Fractions

   Using the pairs $$ℒ\{1\}=\frac{1}{s},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2},$$ find ${ℒ}^{-1}\left\{{\displaystyle \frac{5s^2+3s+16}{s(s^2+4)}}\right\}$.

5. Q5Transforms of Derivatives and Initial Value Problems

   Here $y=y(t)$. Using $ℒ\{y^{\prime }\}=sY(s)-y(0)$, $ℒ\{y^{″}\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)$ and $ℒ\{e^{at}\}={\displaystyle \frac{1}{s-a}}$, solve the initial value problem $$y^{″}-y^{\prime }-6y=0,y(0)=1,y^{\prime }(0)=8$$ by the Laplace transform.

6. Q6Transforms of Derivatives and Initial Value Problems

   Here $y=y(t)$. Using $ℒ\{y^{″}\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)$ and $$ℒ\{t\}=\frac{1}{s^2},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2},$$ solve $y^{″}+4y=8t,y(0)=1,y^{\prime }(0)=0$ by the Laplace transform.

7. Q7Translation on the s-Axis

   The first translation theorem says: if $ℒ\{f(t)\}=F(s)$, then $ℒ\{e^{at}f(t)\}=F(s-a)$. Using it together with $$ℒ\{t^n\}=\frac{n!}{s^{n+1}},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2},$$

   1. find $ℒ\{t^2{e}^{-3t}\}$;
   2. find ${ℒ}^{-1}\left\{{\displaystyle \frac{s+5}{s^2+4s+13}}\right\}$.
8. Q8The Unit Step Function and Translation on the t-Axis

   The unit step function is $u(t-a)=0$ for $t<a$ and $1$ for $t\ge a$, where $a\ge 0$, and $ℒ\{u(t-a)\}={\displaystyle \frac{e^{-as}}{s}}$. Write $$f(t)=\{\begin{array}{cc}2,\hfill & 0\le t<3,\hfill \\ -1,\hfill & 3\le t<5,\hfill \\ 0,\hfill & t\ge 5,\hfill \end{array}$$ as a combination of unit step functions, and find $ℒ\{f(t)\}$.

9. Q9The Unit Step Function and Translation on the t-Axis

   The second translation theorem says: if $ℒ\{f(t)\}=F(s)$ and $a\ge 0$, then $ℒ\{f(t-a)u(t-a)\}=e^{-as}F(s)$. Using it with $ℒ\{1\}={\displaystyle \frac{1}{s}}$, $ℒ\{t^2\}={\displaystyle \frac{2}{s^3}}$ and $ℒ\{e^{ct}\}={\displaystyle \frac{1}{s-c}}$,

   1. find $ℒ\{(t-2{)}^{2}u(t-2)\}$;
   2. find ${ℒ}^{-1}\left\{{\displaystyle \frac{3e^{-4s}}{s(s+3)}}\right\}$, and write the answer piecewise.
10. Q10Initial Value Problems with Piecewise and Periodic Forcing

    Here $y=y(t)$ and $u(t-a)$ is the unit step ($0$ for $t<a$, $1$ for $t\ge a$). Use $ℒ\{y^{″}\}=s^{2}Y-sy(0)-y^{\prime }(0)$, $ℒ\{f(t-a)u(t-a)\}=e^{-as}F(s)$ where $F=ℒ\{f\}$, and $$ℒ\{1\}=\frac{1}{s},ℒ\{\sin bt\}=\frac{b}{s^2+b^2},ℒ\{\cos bt\}=\frac{s}{s^2+b^2}$$ to solve $$y^{″}+4y=8u(t-\pi ),y(0)=0,y^{\prime }(0)=2.$$ Give the solution piecewise.

11. Q11Initial Value Problems with Piecewise and Periodic Forcing

    Here $y=y(t)$. If $f$ is periodic with period $T$, then $ℒ\{f(t)\}={\displaystyle \frac{1}{1-e^{-Ts}}}{\displaystyle {\int }_0^T{e}^{-st}f(t)dt}$. Let $f$ be the square wave of period $2$ with $f(t)=1$ for $0\le t<1$ and $f(t)=0$ for $1\le t<2$.

    1. Find $ℒ\{f(t)\}$ and simplify it.
    2. Using $ℒ\{y^{\prime }\}=sY-y(0)$, find the transform $Y(s)$ of the solution of $y^{\prime }+2y=f(t),y(0)=0$. Do not invert it.
12. Q12The Convolution Theorem

    The convolution of $f$ and $g$ is $(f*g)(t)={\displaystyle {\int }_0^{t}f(\tau )g(t-\tau )d\tau }$, and the convolution theorem says $ℒ\{(f*g)(t)\}=F(s)G(s)$.

    1. Compute $t*e^t$ from the definition.
    2. Check your answer by finding its inverse transform another way: invert $ℒ\{t\}·ℒ\{e^t\}={\displaystyle \frac{1}{s^2(s-1)}}$ by partial fractions, using $ℒ\{1\}={\displaystyle \frac{1}{s}}$, $ℒ\{t\}={\displaystyle \frac{1}{s^2}}$, $ℒ\{e^t\}={\displaystyle \frac{1}{s-1}}$.
13. Q13The Dirac Delta Function and Impulsive Forcing

    Here $y=y(t)$, $\delta (t-a)$ is the Dirac delta at $a>0$, with $ℒ\{\delta (t-a)\}=e^{-as}$, and $u(t-a)$ is the unit step ($0$ for $t<a$, $1$ for $t\ge a$). Using $ℒ\{y^{\prime }\}=sY-y(0)$, $ℒ\{e^{ct}\}={\displaystyle \frac{1}{s-c}}$ and $ℒ\{f(t-a)u(t-a)\}=e^{-as}ℒ\{f(t)\}$, solve $$y^{\prime }+3y=2\delta (t-1),y(0)=4,$$ and state how the impulse changes $y$ at $t=1$.

14. Q14The Dirac Delta Function and Impulsive Forcing

    Here $y=y(t)$, $\delta (t-a)$ is the Dirac delta at $a>0$, with $ℒ\{\delta (t-a)\}=e^{-as}$, and $u(t-a)$ is the unit step. Using $ℒ\{y^{″}\}=s^{2}Y-sy(0)-y^{\prime }(0)$, $ℒ\{f(t-a)u(t-a)\}=e^{-as}ℒ\{f(t)\}$, $ℒ\{e^{ct}\cos t\}={\displaystyle \frac{s-c}{(s-c{)}^2+1}}$ and $ℒ\{e^{ct}\sin t\}={\displaystyle \frac{1}{(s-c{)}^2+1}}$, solve $$y^{″}+2y^{\prime }+2y=\delta (t-2),y(0)=1,y^{\prime }(0)=-1.$$

15. Q15Synthesis — drawing on several topics in this unit

    Here $y=y(t)$ and $u(t-a)$ is the unit step ($0$ for $t<a$, $1$ for $t\ge a$). Using $ℒ\{y^{\prime }\}=sY-y(0)$, $ℒ\{y^{″}\}=s^{2}Y-sy(0)-y^{\prime }(0)$, $ℒ\{1\}={\displaystyle \frac{1}{s}}$, $ℒ\{e^{ct}\}={\displaystyle \frac{1}{s-c}}$ and $ℒ\{f(t-a)u(t-a)\}=e^{-as}ℒ\{f(t)\}$, solve $$y^{″}+3y^{\prime }+2y=2u(t-1),y(0)=0,y^{\prime }(0)=1,$$ and find ${\displaystyle {lim}_{t\to \infty }y(t)}$.

The 10 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The transform from its definition

The Laplace transform of f is an improper integral, ℒ{f} = F(s) = ∫₀^∞ e^(−st) f(t) dt, and it exists only for the values of s that make that integral converge. Q1 asks for it from the definition, for a function given in two pieces, and asks for the values of s as well as the formula. Both halves are marked.

Q2 is the table and the fact that makes the table enough: the transform is linear, so a sum of constant multiples transforms term by term. Part (a) is already in that form. Parts (b) and (c) are not, and the work is to rewrite them until they are — expand the square in (b), and in (c) use a double-angle identity to replace the square of a cosine by something linear in cosines. There is no table entry for a product or a square, so a function that is not already a sum of table entries has to be made into one first.

Inverse transforms and partial fractions

Going back from F(s) to f(t) is reading the table in reverse, and almost all of the work is making F(s) look like a table entry. Q3 has two shapes. In part (a) each term already matches a pair except for its constant, and the constant is adjusted: the pair for tⁿ carries n! on top, so a different numerator means multiplying and dividing by what is missing. In part (b) the numerator has an s term and a constant term over s² + b².

Q4 needs partial fractions first. The denominator is s times an irreducible quadratic, so the decomposition has a constant over s and a linear term over the quadratic. Find the constants by clearing denominators and comparing coefficients (or substituting a convenient s), then invert each piece exactly as in Q3.

Transforms of derivatives and initial value problems

The reason the transform belongs in this course is one formula: the transform of y′ is sY(s) − y(0), and of y″ is s²Y(s) − s y(0) − y′(0). The initial conditions go in at the start, not at the end — there is no general solution with arbitrary constants to fit afterwards.

Q5 is a homogeneous equation: after step 2 the only thing on the right is what the initial conditions produced, and the denominator is a quadratic in s to factor. Q6 has a forcing term, so Y(s) has two parts, one from the forcing and one from the initial conditions, and the forcing part has a repeated factor of s in the denominator as well as s² + 4.

Translation on the s-axis

Multiplying a function by eᵃᵗ shifts its transform: F(s) becomes F(s − a). Q7 uses the theorem in both directions. Part (a) is forwards — find the transform of the power alone, then replace every s by s − a. Part (b) is backwards, and needs a recognisable shift first.

The unit step and translation on the t-axis

The unit step u(t − a) is 0 before a and 1 from a on. It is a switch, and a piecewise function is a sum of switches. Q8 asks you to write a three-piece function that way and then transform it.

The second translation theorem says that delaying a function by a — writing f(t − a) u(t − a) — multiplies its transform by e^(−as). Q9(a) is that statement used forwards: identify what f is once the delay is taken out. Q9(b) is it used backwards, and the exponential is a label, not part of the algebra.

Piecewise and periodic forcing

Q10 is the reason the step function is in this chapter: a forcing term that switches on at t = π, which a trial solution could only handle interval by interval. Transformed, it is one equation. Y(s) splits into the part from the initial conditions and the part from the switched forcing; the second carries e^(−πs), so invert it with the tree above — partial fractions on the rest, then delay.

Q11 gives the formula for a periodic function: the integral over one period, divided by 1 − e^(−Ts). Part (a) needs only the part of the period where the square wave is non-zero, since the rest contributes nothing to the integral; then look for a factorisation of the denominator that cancels with the numerator. Part (b) transforms a first-order equation with that forcing and stops at Y(s) — the question says not to invert, so the answer is a transform.

The convolution theorem

The transform of a product is not the product of the transforms. What does have a product of transforms as its transform is the convolution (f ∗ g)(t) = ∫₀ᵗ f(τ) g(t − τ) dτ. Q12 computes one convolution two ways, and the two answers must agree.

The Dirac delta and impulsive forcing

δ(t − a) models a blow delivered all at once at time a: its transform is e^(−as), which is what you get from the step-function formula if you imagine a very tall, very short pulse of total area 1. In an equation it behaves like any other right-hand side with an exponential label — transform, solve for Y, invert with the delay.

Q13 is first order. Its solution is one formula before t = 1 and another after, joined by a step, and the question asks what the impulse does to y at that moment.

Q14 is second order, with a damped left-hand side, so the denominator of Y(s) is a quadratic you complete the square in — Q7's technique. The table pairs printed with the question are exactly the shifted cosine and sine you will need. The impulse term carries e^(−2s), so it is inverted with the delay; the initial-condition terms are not.

The synthesis question

Q15 puts the chapter together: a second-order equation with a switched constant forcing and non-zero initial slope. It is Q5's four steps with Q10's delay. The part from the switched forcing has three factors in its denominator — one from the step, two from the characteristic polynomial — so its partial fractions have three terms before you delay them.

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Getting the most out of it

Keep the table in front of you, and use only its entries

Every question prints the pairs it expects. Work from those. The whole skill of this chapter is rewriting an expression until each piece matches an entry exactly — including the constant on top — and adjusting by multiplying and dividing, never by changing a denominator.

Write the initial values in at step one

Substitute y(0) and y′(0) into the derivative formulas as soon as you transform. A Laplace solution has no arbitrary constants; if you find yourself fitting one at the end, something went wrong earlier.

Set exponentials aside before partial fractions

A factor e^(−as) is never part of a partial-fraction decomposition. Pull it out, decompose and invert what is left, then apply the delay to the whole of the result. Mark which terms carry a delay before you start, so none is inverted as if it did not.

Check the answer in the equation you started from

Evaluate your solution at t = 0 to check the initial values, and on each interval of a piecewise answer check that it satisfies the equation. For Q5 and Q6, the characteristic-equation methods from the earlier sets give an independent second answer.

Common questions