Practice / Review Problems — MATH 2410

Solution:

By linearity, we can transform each term separately and add the results, so there is no need to touch the definition at all here — we just need three entries from the transform table.

For the constant term, \(\mathcal{L}\{k\}=\dfrac{k}{s}\), so \(\mathcal{L}\{6\}=\dfrac{6}{s}\). For the power of \(t\), \(\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}\) with \(n=3\), so \(\mathcal{L}\{t^3\}=\dfrac{3!}{s^4}=\dfrac{6}{s^4}\), and hence \(\mathcal{L}\{4t^3\}=\dfrac{24}{s^4}\). For the exponential term, \(\mathcal{L}\{e^{at}\}=\dfrac{1}{s-a}\) with \(a=5\), so \(\mathcal{L}\{2e^{5t}\}=\dfrac{2}{s-5}\).

Adding these three pieces together (with the correct signs) gives

\[ \boxed{\mathcal{L}\{6-4t^3+2e^{5t}\} = \frac{6}{s}-\frac{24}{s^4}+\frac{2}{s-5}.} \]

Solution:

"Directly from the definition" means we go back to the defining improper integral rather than quoting the table entry. Starting from the definition,

\[ \mathcal{L}\{e^{2t}\} = \int_0^\infty e^{-st}e^{2t}\,dt. \]

Since the two exponentials share the same base, we can combine them into a single exponential before integrating:

\[ \mathcal{L}\{e^{2t}\} = \int_0^\infty e^{-(s-2)t}\,dt. \]

This is now the same basic exponential integral that appears throughout Section 7.1, so we antidifferentiate and evaluate the (improper) limits of integration:

\[ \int_0^\infty e^{-(s-2)t}\,dt = \lim_{b\to\infty}\left[-\frac{1}{s-2}e^{-(s-2)t}\right]_0^b = \lim_{b\to\infty}\left(-\frac{e^{-(s-2)b}}{s-2}+\frac{1}{s-2}\right). \]

Provided \(s-2>0\), i.e. \(s>2\), the term \(e^{-(s-2)b}\to0\) as \(b\to\infty\), and the limit reduces to just the second term. So

\[ \boxed{\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}, \qquad s>2.} \]

Solution:

By linearity, \(\mathcal{L}\{2t\}=2\,\mathcal{L}\{t\}\), so it suffices to evaluate \(\mathcal{L}\{t\}\) from the definition — this mirrors the \(\mathcal{L}\{t\}\) example already worked out in the Section 7.1 notes.

\[ \mathcal{L}\{t\} = \int_0^\infty te^{-st}\,dt. \]

Step 1 (Integrate by parts).

Since the integrand is a product of \(t\) and \(e^{-st}\), we use integration by parts with \(u=t\), \(dv=e^{-st}\,dt\), so \(du=dt\), \(v=-\frac{1}{s}e^{-st}\):

\[ \int te^{-st}\,dt = -\frac{t}{s}e^{-st} + \frac{1}{s}\int e^{-st}\,dt = -\frac{t}{s}e^{-st} - \frac{1}{s^2}e^{-st}. \]

Step 2 (Evaluate the improper integral).

\[ \int_0^\infty te^{-st}\,dt = \lim_{b\to\infty}\left(-\frac{b}{s}e^{-sb}-\frac{1}{s^2}e^{-sb}\right) - \left(0-\frac{1}{s^2}\right). \]

For \(s>0\), both \(be^{-sb}\) and \(e^{-sb}\) vanish as \(b\to\infty\) (the exponential decay beats the linear growth of \(b\)), so the limit is simply \(\dfrac{1}{s^2}\). Thus \(\mathcal{L}\{t\}=\dfrac{1}{s^2}\), and multiplying by \(2\) gives

\[ \boxed{\mathcal{L}\{2t\} = \frac{2}{s^2}, \qquad s>0.} \]

Solution:

Let \(I=\displaystyle\int_0^\infty e^{-st}\cos(kt)\,dt\). Because differentiating \(\cos\) or \(\sin\) never eliminates the trig function, a single integration by parts won't finish the job — instead, applying it twice will produce the original integral again, and we can then solve algebraically for it (exactly as in the \(\sin t\) example in the notes).

Step 1 (First integration by parts).

Let \(u=e^{-st}\), \(dv=\cos(kt)\,dt\), so \(du=-se^{-st}\,dt\), \(v=\frac{1}{k}\sin(kt)\):

\[ I = \left[\frac{1}{k}e^{-st}\sin(kt)\right]_0^\infty + \frac{s}{k}\int_0^\infty e^{-st}\sin(kt)\,dt. \]

The boundary term vanishes at both ends: at \(t=0\), \(\sin(0)=0\); as \(t\to\infty\), \(e^{-st}\to0\) (for \(s>0\)) while \(\sin(kt)\) stays bounded. So \(I=\dfrac{s}{k}J\), where \(J=\displaystyle\int_0^\infty e^{-st}\sin(kt)\,dt\).

Step 2 (Second integration by parts).

Now apply the same idea to \(J\), with \(u=e^{-st}\), \(dv=\sin(kt)\,dt\), so \(du=-se^{-st}\,dt\), \(v=-\frac{1}{k}\cos(kt)\):

\[ J = \left[-\frac{1}{k}e^{-st}\cos(kt)\right]_0^\infty - \frac{s}{k}\int_0^\infty e^{-st}\cos(kt)\,dt. \]

At \(t\to\infty\) this boundary term again vanishes for \(s>0\); at \(t=0\) it equals \(-\frac{1}{k}(1)(1)=-\frac{1}{k}\). So the bracket contributes \(0-\left(-\frac1k\right)=\frac1k\), and

\[ J = \frac{1}{k} - \frac{s}{k}I. \]

Step 3 (Substitute back and solve for \(I\)).

Substituting this expression for \(J\) into \(I=\frac{s}{k}J\) brings back the original unknown \(I\):

\[ I = \frac{s}{k}\left(\frac{1}{k}-\frac{s}{k}I\right) = \frac{s}{k^2}-\frac{s^2}{k^2}I. \]

Moving the \(I\)-term on the right to the left side and factoring,

\[ I\left(1+\frac{s^2}{k^2}\right) = \frac{s}{k^2} \quad\Longrightarrow\quad I\cdot\frac{k^2+s^2}{k^2}=\frac{s}{k^2} \quad\Longrightarrow\quad I = \frac{s}{s^2+k^2}. \]

So

\[ \boxed{\mathcal{L}\{\cos(kt)\} = \frac{s}{s^2+k^2}, \qquad s>0.} \]

Solution:

By definition, \(\mathcal{L}\{e^{at}\}=\displaystyle\int_0^\infty e^{-st}e^{at}\,dt =\int_0^\infty e^{-(s-a)t}\,dt\), which is exactly the same convergence question as the \(\mathcal{L}\{1\}\) computation in the notes, just with \(s\) replaced by \(s-a\).

Evaluating the antiderivative and taking the limit as the upper bound \(b\to\infty\) gives a term \(e^{-(s-a)b}\). This term tends to \(0\) exactly when the exponent \(-(s-a)b\) tends to \(-\infty\), which happens precisely when \(s-a>0\), i.e. \(s>a\). In that case the improper integral converges to the finite value \(\dfrac{1}{s-a}\).

If instead \(s\le a\), then \(s-a\le0\), so \(-(s-a)\ge0\) and \(e^{-(s-a)t}\) either grows without bound as \(t\to\infty\) (if \(s\lt a\)) or stays constantly equal to \(1\) forever (if \(s=a\)); either way, \(\int_0^b e^{-(s-a)t}\,dt\) does not settle down to a finite number as \(b\to\infty\) — the improper integral diverges, and \(\mathcal{L}\{e^{at}\}\) simply does not exist there. This is why every Laplace transform formula must be quoted together with the range of \(s\) for which it is actually valid: the formula itself may look fine algebraically for any \(s\), but the integral that defines it only converges on part of that range.

Solution:

The denominator \(s^2+16\) already matches the form \(s^2+k^2\) with \(k=4\), so once we split the numerator into two pieces, both terms will match table entries directly — no partial fractions needed here. Splitting,

\[ \frac{7s+12}{s^2+16} = 7\cdot\frac{s}{s^2+16} + 12\cdot\frac{1}{s^2+16}. \]

Using \(\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+k^2}\right\}=\cos(kt)\) and \(\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+k^2}\right\}=\dfrac{1}{k}\sin(kt)\) with \(k=4\), the first term gives \(7\cos(4t)\), and the second gives \(12\cdot\frac14\sin(4t)=3\sin(4t)\). So

\[ \boxed{\mathcal{L}^{-1}\left\{\frac{7s+12}{s^2+16}\right\} = 7\cos(4t)+3\sin(4t).} \]

Solution:

Before reaching for partial fractions, it is always worth checking whether the expression simplifies first. Factoring an \(s\) out of the denominator,

\[ s^3+4s = s(s^2+4), \]

so the whole expression collapses by cancelling the common factor of \(s\):

\[ \frac{s}{s^3+4s} = \frac{s}{s(s^2+4)} = \frac{1}{s^2+4}. \]

This is now a direct table match, \(\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+k^2}\right\}=\dfrac1k\sin(kt)\) with \(k=2\), so

\[ \boxed{\mathcal{L}^{-1}\left\{\frac{s}{s^3+4s}\right\} = \frac{1}{2}\sin(2t).} \]

Solution:

Step 1 (Factor the denominator).

\[ s^2+3s+2=(s+1)(s+2). \]

Step 2 (Set up and solve the partial fraction decomposition).

Since the two roots are distinct and real, we decompose into two simple fractions:

\[ \frac{s-2}{(s+1)(s+2)} = \frac{A}{s+1}+\frac{B}{s+2} \quad\Longrightarrow\quad s-2=A(s+2)+B(s+1). \]

Setting \(s=-1\): \(-3=A(1)\Rightarrow A=-3\). Setting \(s=-2\): \(-4=B(-1)\Rightarrow B=4\).

Step 3 (Take the inverse transform term by term).

Now that the quotient is rewritten as a sum, linearity of \(\mathcal{L}^{-1}\) lets us transform each piece separately using \(\mathcal{L}^{-1}\left\{\dfrac1{s-a}\right\}=e^{at}\):

\[ \boxed{\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+3s+2}\right\} = -3e^{-t}+4e^{-2t}.} \]

Solution:

This is the same partial-fraction technique as before, just with a third distinct linear factor — the extra factor means an extra unknown constant to pin down, but the "plug in each root" trick still works one root at a time.

Step 1 (Set up the decomposition).

\[ \frac{4s^2+9s-4}{s(s-1)(s+2)} = \frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+2} \quad\Longrightarrow\quad 4s^2+9s-4 = A(s-1)(s+2)+Bs(s+2)+Cs(s-1). \]

Step 2 (Solve for each constant by plugging in the roots).

Each choice of \(s\) equal to one of the roots makes two of the three terms on the right vanish, isolating a single constant:

\[ s=0:\ -4=A(-1)(2)=-2A \Rightarrow A=2. \] \[ s=1:\ 4+9-4=9=B(1)(3)=3B \Rightarrow B=3. \] \[ s=-2:\ 16-18-4=-6=C(-2)(-3)=6C \Rightarrow C=-1. \]

Step 3 (Take the inverse transform term by term).

With the decomposition \(\dfrac{2}{s}+\dfrac{3}{s-1}-\dfrac{1}{s+2}\) in hand, each piece is a direct table entry:

\[ \boxed{\mathcal{L}^{-1}\left\{\frac{4s^2+9s-4}{s(s-1)(s+2)}\right\} = 2+3e^{t}-e^{-2t}.} \]

Solution:

Step 1 (Transform both sides).

Writing \(F(s)=\mathcal{L}\{y(t)\}\) and using the derivative formula with the given (zero) initial conditions, \(\mathcal{L}\{y''\}=s^2F(s)-sy(0)-y'(0)=s^2F(s)\). On the right, \(\mathcal{L}\{10t+7\}=\dfrac{10}{s^2}+\dfrac{7}{s}\). So

\[ s^2F(s)+9F(s) = \frac{10}{s^2}+\frac{7}{s}. \]

Step 2 (Solve for \(F(s)\)).

\[ (s^2+9)F(s) = \frac{10+7s}{s^2} \quad\Longrightarrow\quad F(s) = \frac{7s+10}{s^2(s^2+9)}. \]

Step 3 (Partial fractions).

Since \(s^2\) is a repeated factor and \(s^2+9\) is irreducible over the reals, the decomposition needs four constants:

\[ \frac{7s+10}{s^2(s^2+9)} = \frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+9} \quad\Longrightarrow\quad 7s+10 = As(s^2+9)+B(s^2+9)+(Cs+D)s^2. \]

Expanding and collecting powers of \(s\) on the right gives \((A+C)s^3+(B+D)s^2+9As+9B\). Matching this to \(0\cdot s^3+0\cdot s^2+7s+10\) term by term: \(9A=7\Rightarrow A=\frac79\); \(9B=10\Rightarrow B=\frac{10}{9}\); \(A+C=0\Rightarrow C=-\frac79\); \(B+D=0\Rightarrow D=-\frac{10}{9}\). So

\[ F(s) = \frac{7}{9}\cdot\frac1s + \frac{10}{9}\cdot\frac{1}{s^2} - \frac{7}{9}\cdot\frac{s}{s^2+9} - \frac{10}{9}\cdot\frac{1}{s^2+9}. \]

Step 4 (Take the inverse transform).

Using \(\mathcal{L}^{-1}\{1/s\}=1\), \(\mathcal{L}^{-1}\{1/s^2\}=t\), \(\mathcal{L}^{-1}\{s/(s^2+9)\}=\cos(3t)\), and \(\mathcal{L}^{-1}\{1/(s^2+9)\}=\frac13\sin(3t)\) (so the last term contributes \(\frac{10}{9}\cdot\frac13=\frac{10}{27}\)),

\[ \boxed{y(t) = \frac{10}{9}t+\frac{7}{9}-\frac{7}{9}\cos(3t)-\frac{10}{27}\sin(3t).} \]

Solution:

Linearity of \(\mathcal{L}^{-1}\) says that \(\mathcal{L}^{-1}\{c_1F(s)+c_2G(s)\} =c_1\mathcal{L}^{-1}\{F(s)\}+c_2\mathcal{L}^{-1}\{G(s)\}\) — it distributes over sums and scalar multiples, and only over sums and scalar multiples. There is no corresponding rule for products or quotients: in general \(\mathcal{L}^{-1}\{F(s)/G(s)\}\neq \mathcal{L}^{-1}\{F(s)\}/\mathcal{L}^{-1}\{G(s)\}\) (transforming a numerator and denominator independently and then dividing the resulting functions of \(t\) has no mathematical justification at all).

So a single quotient like \(\dfrac{6s+2}{s^2-1}\) cannot be inverse-transformed directly unless it happens to already match a table entry. Partial fraction decomposition rewrites this one quotient as a genuine sum of simpler quotients, \(\dfrac{4}{s-1}+\dfrac{2}{s+1}\), each of which matches a table entry on its own. Only once the expression is in this additive form does linearity legally apply, letting us transform each piece separately and add the results. In short: partial fractions doesn't make the inverse transform "linear" — it restates the problem in the one form (a sum) where linearity was already applicable.

Solution:

The function \(te^{3t}\) is exactly \(e^{at}f(t)\) with \(a=3\) and \(f(t)=t\), so this is a direct application of the First Translation Theorem, \(\mathcal{L}\{e^{at}f(t)\}=F(s-a)\), where \(F(s)=\mathcal{L}\{f(t)\}\).

Here \(F(s)=\mathcal{L}\{t\}=\dfrac{1}{s^2}\). Replacing every \(s\) in \(F(s)\) with \(s-3\) (the "shift" the theorem describes) gives

\[ \boxed{\mathcal{L}\{te^{3t}\} = \frac{1}{(s-3)^2}.} \]

Solution:

Whenever a piecewise function is really "one base formula, plus an extra piece that switches on at some point," the unit step function lets us combine it into a single expression instead of splitting the transform integral in two. Here the base formula \(t^2\) runs the whole time, and an extra constant \(1\) switches on at \(t=3\):

\[ f(t) = t^2 + \mathscr{U}(t-3). \]

(Check: for \(t<3\), \(\mathscr U(t-3)=0\) and \(f(t)=t^2\); for \(t\ge3\), \(\mathscr U(t-3)=1\) and \(f(t)=t^2+1\) — matching the given definition.)

Now transform each piece: \(\mathcal{L}\{t^2\}=\dfrac{2}{s^3}\) from the table, and \(\mathcal{L}\{\mathscr U(t-3)\}=\dfrac{e^{-3s}}{s}\) (the special case of the Second Translation Theorem where the "delayed function" is just the constant \(1\)). Adding these,

\[ \boxed{\mathcal{L}\{f(t)\} = \frac{2}{s^3}+\frac{e^{-3s}}{s}.} \]

Solution:

First split the function using linearity, so that the First Translation Theorem can be applied to each piece separately:

\[ (t^2+2t)e^t = t^2e^t + 2te^t. \]

From the table, \(\mathcal{L}\{t^2\}=\dfrac{2}{s^3}\) and \(\mathcal{L}\{t\}=\dfrac{1}{s^2}\). Applying the shift \(s\to s-1\) (since \(a=1\) here) to each:

\[ \mathcal{L}\{t^2e^t\} = \frac{2}{(s-1)^3}, \qquad \mathcal{L}\{2te^t\} = \frac{2}{(s-1)^2}. \]

Adding the two pieces together,

\[ \boxed{\mathcal{L}\{(t^2+2t)e^{t}\} = \frac{2}{(s-1)^3}+\frac{2}{(s-1)^2}.} \]

Solution:

The Second Translation Theorem, \(\mathcal{L}\{f(t-a)\mathscr U(t-a)\}=e^{-as}F(s)\), applies to a function of \(t-a\) multiplying \(\mathscr U(t-a)\) — but here the factor multiplying \(\mathscr U(t-2)\) is \(t-1\), not \(t-2\). So the first step is purely algebraic: rewrite \(t-1\) in terms of \((t-2)\).

Step 1 (Rewrite in terms of \(t-2\)).

\[ t-1 = (t-2)+1 \quad\Longrightarrow\quad (t-1)\mathscr U(t-2) = (t-2)\mathscr U(t-2) + \mathscr U(t-2). \]

Step 2 (Apply the Second Translation Theorem to each piece).

The first piece has \(f(t)=t\) shifted by \(a=2\), so \(\mathcal{L}\{(t-2)\mathscr U(t-2)\}=e^{-2s}\mathcal{L}\{t\}=e^{-2s}\cdot\dfrac1{s^2}\). The second piece is the constant \(1\) shifted by \(a=2\), so \(\mathcal{L}\{\mathscr U(t-2)\}=e^{-2s}\cdot\dfrac1s\).

Adding the two pieces,

\[ \boxed{\mathcal{L}\{(t-1)\mathscr{U}(t-2)\} = e^{-2s}\left(\frac{1}{s^2}+\frac{1}{s}\right).} \]

Solution:

Step 1 (Transform both sides).

With \(F(s)=\mathcal{L}\{y(t)\}\) and the given initial conditions, \(\mathcal{L}\{y''\}=s^2F(s)-s\) and \(\mathcal{L}\{y'\}=sF(s)-1\). On the right, the forcing term needs the First Translation Theorem: since \(\mathcal{L}\{\cos t\}=\dfrac{s}{s^2+1}\), shifting by \(a=1\) gives \(\mathcal{L}\{e^t\cos t\}=\dfrac{s-1}{(s-1)^2+1}\). So

\[ \big(s^2F(s)-s\big) - 5\big(sF(s)-1\big)+6F(s) = \frac{s-1}{(s-1)^2+1}. \]

Step 2 (Solve for \(F(s)\)).

\[ (s^2-5s+6)F(s) = \frac{s-1}{(s-1)^2+1}+s-5 \quad\Longrightarrow\quad F(s) = \underbrace{\frac{s-1}{(s-2)(s-3)\big[(s-1)^2+1\big]}}_{\text{piece 1}} + \underbrace{\frac{s-5}{(s-2)(s-3)}}_{\text{piece 2}}. \]

Step 3 (Partial fractions on piece 2).

\[ \frac{s-5}{(s-2)(s-3)} = \frac{A}{s-2}+\frac{B}{s-3} \quad\Longrightarrow\quad s-5=A(s-3)+B(s-2). \]

Setting \(s=2\): \(-3=-A\Rightarrow A=3\). Setting \(s=3\): \(-2=B\Rightarrow B=-2\). So piece 2 is \(\dfrac{3}{s-2}-\dfrac{2}{s-3}\).

Step 4 (Partial fractions on piece 1).

Since \((s-1)^2+1\) is an irreducible quadratic, piece 1 needs the form

\[ \frac{s-1}{(s-2)(s-3)\big[(s-1)^2+1\big]} = \frac{A'}{s-2}+\frac{B'}{s-3}+\frac{Cs+D}{(s-1)^2+1}. \]

Plugging in \(s=2\) and \(s=3\) isolates \(A'\) and \(B'\) just as before (the quadratic factor doesn't vanish at either root, so it just comes along for the ride):

\[ s=2:\ 1 = A'(-1)\big[(1)^2+1\big] = -2A' \Rightarrow A'=-\tfrac12. \] \[ s=3:\ 2 = B'(1)\big[(2)^2+1\big] = 5B' \Rightarrow B'=\tfrac25. \]

Matching the remaining coefficients (comparing the \(s^3\), \(s^2\), and constant terms after clearing denominators) gives \(C=\tfrac{1}{10}\), \(D=-\tfrac25\) — so piece 1 is \(\dfrac{-1/2}{s-2}+\dfrac{2/5}{s-3}+\dfrac{(1/10)s-2/5}{(s-1)^2+1}\).

Step 5 (Combine both pieces).

Adding piece 1 and piece 2 term by term: the \(1/(s-2)\) coefficients give \(3-\tfrac12=\tfrac52\); the \(1/(s-3)\) coefficients give \(-2+\tfrac25=-\tfrac85\). So

\[ F(s) = \frac{5/2}{s-2} - \frac{8/5}{s-3} + \frac{(1/10)s-2/5}{(s-1)^2+1}. \]

Step 6 (Take the inverse transform).

For the last term, write the numerator in terms of \((s-1)\) so it matches the shifted table entries: \(\frac{1}{10}s-\frac25 = \frac{1}{10}(s-1)-\frac{3}{10}\). Then, using \(\mathcal{L}^{-1}\left\{\dfrac{s-1}{(s-1)^2+1}\right\}=e^t\cos t\) and \(\mathcal{L}^{-1}\left\{\dfrac{1}{(s-1)^2+1}\right\}=e^t\sin t\) (both from the First Translation Theorem in reverse),

\[ \boxed{y(t) = \frac{5}{2}e^{2t} - \frac{8}{5}e^{3t} + \frac{1}{10}e^{t}\cos t - \frac{3}{10}e^{t}\sin t.} \]

(As a check, \(y(0)=\frac52-\frac85+\frac1{10}=\frac{25-16+1}{10}=1\), matching the initial condition.)

Solution:

Step 1 (Transform both sides).

With \(y(0)=0\), \(\mathcal{L}\{y'\}=sF(s)\). On the right, \(\mathcal{L}\{\mathscr U(t-2)\}=\dfrac{e^{-2s}}{s}\), so

\[ sF(s)+3F(s) = 6\cdot\frac{e^{-2s}}{s}. \]

Step 2 (Solve for \(F(s)\)).

\[ F(s) = \frac{6e^{-2s}}{s(s+3)}. \]

Step 3 (Partial fractions on the non-shifted part).

Setting aside the \(e^{-2s}\) factor for a moment, decompose \(\dfrac{6}{s(s+3)}=\dfrac As+\dfrac{B}{s+3}\): \(6=A(s+3)+Bs\). At \(s=0\): \(A=2\). At \(s=-3\): \(-3B=6\Rightarrow B=-2\). So

\[ F(s) = e^{-2s}\left(\frac{2}{s}-\frac{2}{s+3}\right). \]

Step 4 (Apply the Second Translation Theorem).

Let \(G(s)=\dfrac2s-\dfrac2{s+3}\), so \(g(t)=\mathcal{L}^{-1}\{G(s)\}=2-2e^{-3t}\). By the Second Translation Theorem, \(\mathcal{L}^{-1}\{e^{-2s}G(s)\}=g(t-2)\mathscr U(t-2)\), i.e. we substitute \(t-2\) in place of \(t\) everywhere in \(g\) and multiply by \(\mathscr U(t-2)\):

\[ \boxed{y(t) = \Big(2-2e^{-3(t-2)}\Big)\mathscr{U}(t-2).} \]

This says the tank/system stays at rest (\(y=0\)) until \(t=2\), exactly matching \(y(0)=0\) and the fact that the forcing term is off until then, and only afterward rises toward the value \(2\).

Solution:

Each theorem takes an operation performed in one domain (\(t\) or \(s\)) and shows that it corresponds to the other kind of operation in the opposite domain — and the two theorems swap which operation goes with which domain.

In the First Translation Theorem, we multiply by an exponential \(e^{at}\) in the \(t\)-domain (before ever transforming), and the effect on the transform side is a shift: every \(s\) in \(F(s)\) gets replaced by \(s-a\). In the Second Translation Theorem, it is exactly the other way around: we shift \(f\) in the \(t\)-domain (delaying it by \(a\) and switching it on with \(\mathscr U(t-a)\)), and the effect on the transform side is to multiply \(F(s)\) by the exponential \(e^{-as}\).

So multiplication-by-an-exponential in one domain always pairs with a shift in the other domain, and which domain plays which role simply flips between the two theorems: "exponential in \(t\) \(\Rightarrow\) shift in \(s\)" versus "shift in \(t\) \(\Rightarrow\) exponential in \(s\)." Recognizing this duality is often the fastest way to tell, at a glance, which theorem a given problem calls for.

Solution:

This is a direct application of the derivative-of-transform theorem with \(n=1\): \(\mathcal{L}\{tf(t)\}=-F'(s)\), where \(F(s)=\mathcal{L}\{f(t)\}\).

Here \(f(t)=\sin(2t)\), so \(F(s)=\dfrac{2}{s^2+4}\). Differentiating with the chain rule,

\[ F'(s) = 2\cdot\frac{-2s}{(s^2+4)^2} = \frac{-4s}{(s^2+4)^2}. \]

Multiplying by \(-1\) as the theorem requires,

\[ \boxed{\mathcal{L}\{t\sin(2t)\} = \frac{4s}{(s^2+4)^2}.} \]

Solution:

This problem stacks two tools from this unit: the First Translation Theorem handles the \(e^{-2t}\) factor, and then the derivative-of-transform theorem handles the extra \(t\).

Step 1 (Apply the First Translation Theorem).

Start from \(\mathcal{L}\{\cos(5t)\}=\dfrac{s}{s^2+25}\). Shifting by \(a=-2\) (i.e. replacing \(s\) with \(s-(-2)=s+2\)) gives the transform of \(e^{-2t}\cos(5t)\) itself:

\[ F(s) = \mathcal{L}\{e^{-2t}\cos(5t)\} = \frac{s+2}{(s+2)^2+25}. \]

Step 2 (Apply the derivative-of-transform theorem).

Now \(\mathcal{L}\{t\cdot[e^{-2t}\cos(5t)]\}=-F'(s)\). Writing \(u=s+2\) so \(F=\dfrac{u}{u^2+25}\) and \(\frac{d}{ds}=\frac{d}{du}\) (since \(du/ds=1\)), the quotient rule gives

\[ F'(s) = \frac{(u^2+25)-u(2u)}{(u^2+25)^2} = \frac{25-u^2}{(u^2+25)^2} = \frac{25-(s+2)^2}{\big((s+2)^2+25\big)^2}. \]

Multiplying by \(-1\),

\[ \boxed{\mathcal{L}\{te^{-2t}\cos(5t)\} = \frac{(s+2)^2-25}{\big((s+2)^2+25\big)^2}.} \]

Solution:

Here \(n=2\) in the derivative-of-transform theorem, \(\mathcal{L}\{t^2f(t)\}=F''(s)\) (recall \((-1)^2=1\)), so this problem takes the differentiation from Problem 19 one step further.

Starting from \(F(s)=\mathcal{L}\{\sin(2t)\}=\dfrac{2}{s^2+4}\), the first derivative (computed the same way as in Problem 19) is

\[ F'(s) = \frac{-4s}{(s^2+4)^2}. \]

Differentiate once more, using the quotient rule with numerator \(-4s\) and denominator \((s^2+4)^2\) (whose derivative is \(2(s^2+4)(2s)=4s(s^2+4)\)):

\[ F''(s) = \frac{-4(s^2+4)^2 - (-4s)\cdot 4s(s^2+4)}{(s^2+4)^4} = \frac{(s^2+4)\big[-4(s^2+4)+16s^2\big]}{(s^2+4)^4} = \frac{12s^2-16}{(s^2+4)^3}. \]

Since the theorem calls for \((-1)^2F''(s)=F''(s)\) with no sign change,

\[ \boxed{\mathcal{L}\{t^2\sin(2t)\} = \frac{12s^2-16}{(s^2+4)^3}.} \]

Solution:

This combines all three tools at once: the First Translation Theorem for the \(e^{2t}\) factor, followed by two derivatives of the resulting transform for the \(t^2\) factor.

Step 1 (Apply the First Translation Theorem).

From \(\mathcal{L}\{\sin(3t)\}=\dfrac{3}{s^2+9}\), shifting by \(a=2\) gives the transform of \(e^{2t}\sin(3t)\):

\[ F(s) = \frac{3}{(s-2)^2+9}. \]

Step 2 (Differentiate twice).

Let \(u=s-2\), so \(F=\dfrac{3}{u^2+9}\) and derivatives in \(s\) equal derivatives in \(u\). First,

\[ F'(s) = 3\cdot\frac{-2u}{(u^2+9)^2} = \frac{-6u}{(u^2+9)^2}. \]

Differentiating again with the quotient rule (numerator \(-6u\), denominator \((u^2+9)^2\), whose derivative is \(4u(u^2+9)\)):

\[ F''(s) = \frac{-6(u^2+9)^2 - (-6u)(4u)(u^2+9)}{(u^2+9)^4} = \frac{(u^2+9)\big[-6(u^2+9)+24u^2\big]}{(u^2+9)^4} = \frac{18u^2-54}{(u^2+9)^3}. \]

Step 3 (Substitute back and apply the sign).

Since \((-1)^2=1\), \(\mathcal{L}\{t^2e^{2t}\sin(3t)\}=F''(s)\); substituting \(u=s-2\) back in,

\[ \boxed{\mathcal{L}\{t^2 e^{2t}\sin(3t)\} = \frac{18(s-2)^2-54}{\big((s-2)^2+9\big)^3}.} \]

Solution:

The sign traces back to a single derivative: since \(F(s)=\displaystyle\int_0^\infty e^{-st}f(t)\,dt\), differentiating under the integral sign with respect to \(s\) brings down one factor of \(\dfrac{\partial}{\partial s}(e^{-st})=-te^{-st}\) — that is, each \(s\)-derivative multiplies the integrand by an extra factor of \(-t\), not \(t\).

Differentiating once gives \(F'(s)=\displaystyle\int_0^\infty(-t)e^{-st}f(t)\,dt =-\mathcal{L}\{tf(t)\}\), so \(\mathcal{L}\{tf(t)\}=-F'(s)\), matching \(n=1\) in the theorem (\((-1)^1=-1\)). Differentiating a second time brings down another factor of \(-t\), so the integrand now carries \((-t)^2=t^2\) — the two minus signs cancel — giving \(F''(s)=\mathcal{L}\{t^2f(t)\}\), matching \(n=2\) (\((-1)^2=1\)).

In general, differentiating \(n\) times brings down \(n\) factors of \(-t\), i.e. a factor of \((-t)^n=(-1)^nt^n\), so \(F^{(n)}(s)=\mathcal{L}\{(-1)^nt^nf(t)\} =(-1)^n\mathcal{L}\{t^nf(t)\}\). Multiplying both sides by \((-1)^n\) and using \(\big((-1)^n\big)^2=1\) gives \(\mathcal{L}\{t^nf(t)\}=(-1)^nF^{(n)}(s)\) — exactly the theorem. The sign is not a separate fact to memorize; it is forced by the chain rule applied to \(e^{-st}\), and it alternates because each differentiation contributes its own factor of \(-1\) that either matches or cancels the ones already there depending on whether \(n\) is odd or even.