Companion to the University-Style Practice Set配套大学风格练习题集
Sections 1 to 7: definition and existence, transform table, inverse transforms, IVPs, step functions and shifting, the Dirac delta, convolution第 1 至 7 节：定义与存在性、变换表、逆变换、初值问题、阶跃函数与平移、狄拉克 delta 函数、卷积CALC IV
Use $\mathcal{L}\{f\}(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt$ to compute: (a) $\mathcal{L}\{e^{3t}\}$; (b) $\mathcal{L}\{\cos 2t\}$; (c) $\mathcal{L}\{5e^{3t}-3\cos 2t\}$ by linearity.利用 $\mathcal{L}\{f\}(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt$ 计算：(a) $\mathcal{L}\{e^{3t}\}$；(b) $\mathcal{L}\{\cos 2t\}$；(c) 用线性性求 $\mathcal{L}\{5e^{3t}-3\cos 2t\}$。
Apply the definition with $f(t)=e^{3t}$: (M1)将定义用于 $f(t)=e^{3t}$：(M1)
$$\mathcal{L}\{e^{3t}\}=\int_{0}^{\infty}e^{-st}e^{3t}\,dt=\int_{0}^{\infty}e^{-(s-3)t}\,dt.$$For $s>3$ this integral converges. (A1) Evaluating:当 $s>3$ 时此积分收敛。(A1) 计算得：
$$\left[\frac{e^{-(s-3)t}}{-(s-3)}\right]_{0}^{\infty}=0-\frac{1}{-(s-3)}=\frac{1}{s-3}.\quad\text{(A1)}$$Let $I=\int_{0}^{\infty}e^{-st}\cos 2t\,dt$. Integrate by parts: $u=\cos 2t$, $dv=e^{-st}dt$, so $du=-2\sin 2t\,dt$, $v=-\tfrac{1}{s}e^{-st}$. (M1)令 $I=\int_{0}^{\infty}e^{-st}\cos 2t\,dt$。分部积分：取 $u=\cos 2t$，$dv=e^{-st}dt$，则 $du=-2\sin 2t\,dt$，$v=-\tfrac{1}{s}e^{-st}$。(M1)
$$I=\left[-\frac{e^{-st}\cos 2t}{s}\right]_{0}^{\infty}-\frac{2}{s}\int_{0}^{\infty}e^{-st}\sin 2t\,dt=\frac{1}{s}-\frac{2}{s}J,$$where $J=\int_{0}^{\infty}e^{-st}\sin 2t\,dt$. Integrate $J$ by parts: $u=\sin 2t$, $dv=e^{-st}dt$: (M1)其中 $J=\int_{0}^{\infty}e^{-st}\sin 2t\,dt$。对 $J$ 再次分部积分：取 $u=\sin 2t$，$dv=e^{-st}dt$：(M1)
$$J=\left[-\frac{e^{-st}\sin 2t}{s}\right]_{0}^{\infty}+\frac{2}{s}\int_{0}^{\infty}e^{-st}\cos 2t\,dt=0+\frac{2}{s}I.$$Substituting back: $I=\tfrac{1}{s}-\tfrac{2}{s}\cdot\tfrac{2}{s}I=\tfrac{1}{s}-\tfrac{4}{s^{2}}I$. Solving for $I$: $I(1+\tfrac{4}{s^{2}})=\tfrac{1}{s}$, so (A1)代回得：$I=\tfrac{1}{s}-\tfrac{2}{s}\cdot\tfrac{2}{s}I=\tfrac{1}{s}-\tfrac{4}{s^{2}}I$。解出 $I$：$I(1+\tfrac{4}{s^{2}})=\tfrac{1}{s}$，因此 (A1)
$$\mathcal{L}\{\cos 2t\}=I=\frac{s}{s^{2}+4},\quad s>0.$$By linearity of the transform (M1): $\mathcal{L}\{5e^{3t}-3\cos 2t\}=5\cdot\dfrac{1}{s-3}-3\cdot\dfrac{s}{s^{2}+4}$. (A1)由变换的线性性 (M1)：$\mathcal{L}\{5e^{3t}-3\cos 2t\}=5\cdot\dfrac{1}{s-3}-3\cdot\dfrac{s}{s^{2}+4}$。(A1)
Apply the first shift theorem $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$: (a) $\mathcal{L}\{e^{-2t}\sin 3t\}$; (b) $\mathcal{L}\{t^{2}e^{4t}\}$; (c) $\mathcal{L}\{e^{t}(2\cos 3t-5\sin 3t)\}$.应用第一平移定理 $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$ 计算：(a) $\mathcal{L}\{e^{-2t}\sin 3t\}$；(b) $\mathcal{L}\{t^{2}e^{4t}\}$；(c) $\mathcal{L}\{e^{t}(2\cos 3t-5\sin 3t)\}$。
Start from the table entry $\mathcal{L}\{\sin 3t\}=\dfrac{3}{s^{2}+9}$, so $F(s)=\dfrac{3}{s^{2}+9}$. (M1) The multiplier is $e^{-2t}$, so $a=-2$. Replace $s$ by $s-(-2)=s+2$ in $F$: (A1)从变换表项 $\mathcal{L}\{\sin 3t\}=\dfrac{3}{s^{2}+9}$ 出发，即 $F(s)=\dfrac{3}{s^{2}+9}$。(M1) 乘数为 $e^{-2t}$，故 $a=-2$。在 $F$ 中将 $s$ 换为 $s-(-2)=s+2$：(A1)
$$\mathcal{L}\{e^{-2t}\sin 3t\}=F(s+2)=\frac{3}{(s+2)^{2}+9}.\quad\text{(A1)}$$Start from $\mathcal{L}\{t^{2}\}=\dfrac{2!}{s^{3}}=\dfrac{2}{s^{3}}$, so $F(s)=\dfrac{2}{s^{3}}$. (M1) With $a=4$, replace $s$ by $s-4$: (A1)从 $\mathcal{L}\{t^{2}\}=\dfrac{2!}{s^{3}}=\dfrac{2}{s^{3}}$ 出发，即 $F(s)=\dfrac{2}{s^{3}}$。(M1) 取 $a=4$，将 $s$ 换为 $s-4$：(A1)
$$\mathcal{L}\{t^{2}e^{4t}\}=\frac{2}{(s-4)^{3}}.\quad\text{(A1)}$$Split by linearity: $\mathcal{L}\{e^{t}(2\cos 3t-5\sin 3t)\}=2\mathcal{L}\{e^{t}\cos 3t\}-5\mathcal{L}\{e^{t}\sin 3t\}$. (M1) Apply the s-shift with $a=1$ to each, starting from $\mathcal{L}\{\cos 3t\}=\tfrac{s}{s^{2}+9}$ and $\mathcal{L}\{\sin 3t\}=\tfrac{3}{s^{2}+9}$: (A1)利用线性性拆分：$\mathcal{L}\{e^{t}(2\cos 3t-5\sin 3t)\}=2\mathcal{L}\{e^{t}\cos 3t\}-5\mathcal{L}\{e^{t}\sin 3t\}$。(M1) 对每项以 $a=1$ 进行 $s$ 平移，从 $\mathcal{L}\{\cos 3t\}=\tfrac{s}{s^{2}+9}$ 和 $\mathcal{L}\{\sin 3t\}=\tfrac{3}{s^{2}+9}$ 出发：(A1)
$$=\frac{2(s-1)}{(s-1)^{2}+9}-\frac{15}{(s-1)^{2}+9}.$$Find $f(t)=\mathcal{L}^{-1}\{F(s)\}$: (a) $F(s)=\dfrac{3s+1}{s^{2}-s-2}$; (b) $F(s)=\dfrac{2s+6}{s^{2}+4s+13}$.求 $f(t)=\mathcal{L}^{-1}\{F(s)\}$：(a) $F(s)=\dfrac{3s+1}{s^{2}-s-2}$；(b) $F(s)=\dfrac{2s+6}{s^{2}+4s+13}$。
Factor the denominator: $s^{2}-s-2=(s-2)(s+1)$. (M1) Write the partial fraction decomposition:对分母因式分解：$s^{2}-s-2=(s-2)(s+1)$。(M1) 写出部分分式分解：
$$\frac{3s+1}{(s-2)(s+1)}=\frac{A}{s-2}+\frac{B}{s+1}.$$Multiply through: $3s+1=A(s+1)+B(s-2)$. Set $s=2$: $7=3A$, so $A=\tfrac{7}{3}$. Set $s=-1$: $-2=-3B$, so $B=\tfrac{2}{3}$. (M1)两边乘以分母：$3s+1=A(s+1)+B(s-2)$。令 $s=2$：$7=3A$，故 $A=\tfrac{7}{3}$。令 $s=-1$：$-2=-3B$，故 $B=\tfrac{2}{3}$。(M1)
Invert term by term using $\mathcal{L}^{-1}\!\{1/(s-a)\}=e^{at}$: (A1)利用 $\mathcal{L}^{-1}\!\{1/(s-a)\}=e^{at}$ 逐项求逆变换：(A1)
$$f(t)=\frac{7}{3}e^{2t}+\frac{2}{3}e^{-t}.$$Complete the square in the denominator: $s^{2}+4s+13=(s+2)^{2}+9$. (M1) Rewrite the numerator to match shifted $\cos$ and $\sin$ patterns:对分母配方：$s^{2}+4s+13=(s+2)^{2}+9$。(M1) 改写分子以匹配平移后的 $\cos$ 和 $\sin$ 形式：
$$\frac{2s+6}{(s+2)^{2}+9}=\frac{2(s+2)+2}{(s+2)^{2}+9}=\frac{2(s+2)}{(s+2)^{2}+9}+\frac{2}{(s+2)^{2}+9}.$$(M1) The first term matches $\mathcal{L}\{e^{-2t}\cos 3t\}$ and the second is $\tfrac{2}{3}\cdot\tfrac{3}{(s+2)^{2}+9}=\tfrac{2}{3}\mathcal{L}\{e^{-2t}\sin 3t\}$. (A1)(M1) 第一项对应 $\mathcal{L}\{e^{-2t}\cos 3t\}$，第二项为 $\tfrac{2}{3}\cdot\tfrac{3}{(s+2)^{2}+9}=\tfrac{2}{3}\mathcal{L}\{e^{-2t}\sin 3t\}$。(A1)
$$f(t)=2e^{-2t}\cos 3t+\frac{2}{3}e^{-2t}\sin 3t.$$(a) $\mathcal{L}\{u_{2}(t)(t-2)^{3}\}$; (b) $\mathcal{L}^{-1}\!\left\{\tfrac{e^{-3s}}{s^{2}+9}\right\}$; (c) $\mathcal{L}\{u_{1}(t)\cdot t^{2}\}$ after rewriting $t^{2}$ as a polynomial in $t-1$.(a) $\mathcal{L}\{u_{2}(t)(t-2)^{3}\}$；(b) $\mathcal{L}^{-1}\!\left\{\tfrac{e^{-3s}}{s^{2}+9}\right\}$；(c) 将 $t^{2}$ 改写为关于 $t-1$ 的多项式后，求 $\mathcal{L}\{u_{1}(t)\cdot t^{2}\}$。
The function is already in the form $u_{c}(t)f(t-c)$ with $c=2$ and $f(t)=t^{3}$. (M1) Since $\mathcal{L}\{t^{3}\}=3!/s^{4}=6/s^{4}$, the theorem gives:函数已是 $u_{c}(t)f(t-c)$ 的形式，其中 $c=2$，$f(t)=t^{3}$。(M1) 由于 $\mathcal{L}\{t^{3}\}=3!/s^{4}=6/s^{4}$，定理给出：
$$\mathcal{L}\{u_{2}(t)(t-2)^{3}\}=e^{-2s}\cdot\frac{6}{s^{4}}.\quad\text{(A1)}$$The factor $e^{-3s}$ signals a delay of $c=3$. The remaining factor is $F(s)=\tfrac{1}{s^{2}+9}$, which inverts to $f(t)=\tfrac{1}{3}\sin 3t$. (M1) By the inverse theorem, replace $t$ by $t-3$ and gate with $u_{3}(t)$:因子 $e^{-3s}$ 表示延迟 $c=3$。剩余因子 $F(s)=\tfrac{1}{s^{2}+9}$ 的逆变换为 $f(t)=\tfrac{1}{3}\sin 3t$。(M1) 由逆定理，将 $t$ 替换为 $t-3$ 并用 $u_{3}(t)$ 门控：
$$\mathcal{L}^{-1}\!\left\{\frac{e^{-3s}}{s^{2}+9}\right\}=\frac{1}{3}u_{3}(t)\sin 3(t-3).\quad\text{(A1)}$$Write $t=(t-1)+1$, so $t^{2}=((t-1)+1)^{2}=(t-1)^{2}+2(t-1)+1$. (M1) Now each term is in the form required by the theorem with $c=1$:令 $t=(t-1)+1$，则 $t^{2}=((t-1)+1)^{2}=(t-1)^{2}+2(t-1)+1$。(M1) 现在每项均满足定理在 $c=1$ 时所要求的形式：
$$\mathcal{L}\{u_{1}(t)t^{2}\}=e^{-s}\!\left(\frac{2}{s^{3}}+\frac{2}{s^{2}}+\frac{1}{s}\right).\quad\text{(A1)}$$Derive $\mathcal{L}\{y'\}=sY(s)-y(0)$ by integration by parts (a), and use it iteratively to derive $\mathcal{L}\{y''\}=s^{2}Y(s)-sy(0)-y'(0)$ (b).用分部积分推导 $\mathcal{L}\{y'\}=sY(s)-y(0)$（(a)），再迭代应用推导 $\mathcal{L}\{y''\}=s^{2}Y(s)-sy(0)-y'(0)$（(b)）。
Start from the definition: $\mathcal{L}\{y'\}=\displaystyle\int_{0}^{\infty}e^{-st}y'(t)\,dt$. Choose $u=e^{-st}$ and $dv=y'(t)\,dt$, so $du=-se^{-st}\,dt$ and $v=y(t)$. (M1) Integration by parts gives:从定义出发：$\mathcal{L}\{y'\}=\displaystyle\int_{0}^{\infty}e^{-st}y'(t)\,dt$。取 $u=e^{-st}$，$dv=y'(t)\,dt$，则 $du=-se^{-st}\,dt$，$v=y(t)$。(M1) 分部积分得：
$$\mathcal{L}\{y'\}=\Big[e^{-st}y(t)\Big]_{0}^{\infty}-\int_{0}^{\infty}(-s)e^{-st}y(t)\,dt.$$The boundary term at $t\to\infty$: since $y$ is of exponential order, $|y(t)|\le Me^{at}$ for some constants $M,a$, so $|e^{-st}y(t)|\le Me^{-(s-a)t}\to 0$ as $t\to\infty$ provided $s>a$. Hence the boundary term at infinity is $0$. (M1) At $t=0$ the term is $e^{0}\cdot y(0)=y(0)$. So:$t\to\infty$ 处的边界项：由于 $y$ 为指数阶，存在常数 $M,a$ 使 $|y(t)|\le Me^{at}$，从而 $|e^{-st}y(t)|\le Me^{-(s-a)t}\to 0$（$t\to\infty$，只要 $s>a$）。故无穷处边界项为 $0$。(M1) 在 $t=0$ 处边界项为 $e^{0}\cdot y(0)=y(0)$。因此：
$$\mathcal{L}\{y'\}=\bigl(0-y(0)\bigr)+s\int_{0}^{\infty}e^{-st}y(t)\,dt=sY(s)-y(0).\quad\text{(A1·A1)}$$Write $y''=(y')'$ and apply the formula from (a) to the function $y'$ in place of $y$. (M1) Setting $f=y'$ in the formula gives $\mathcal{L}\{(y')'\}=s\mathcal{L}\{y'\}-y'(0)$. (M1)将 $y''=(y')'$，对函数 $y'$ 代替 $y$ 应用 (a) 的公式。(M1) 在公式中令 $f=y'$，得 $\mathcal{L}\{(y')'\}=s\mathcal{L}\{y'\}-y'(0)$。(M1)
Substitute the result of (a) for $\mathcal{L}\{y'\}$: (A1)将 (a) 的结果代入 $\mathcal{L}\{y'\}$：(A1)
$$\mathcal{L}\{y''\}=s\bigl(sY(s)-y(0)\bigr)-y'(0)=s^{2}Y(s)-sy(0)-y'(0).\quad\text{(A1)}$$Prove the first shift theorem (a) and the second shift theorem (b), then apply one in (c).证明第一平移定理 (a) 和第二平移定理 (b)，并在 (c) 中应用其中一个。
By definition: (M1)由定义：(M1)
$$\mathcal{L}\{e^{at}f(t)\}=\int_{0}^{\infty}e^{-st}e^{at}f(t)\,dt=\int_{0}^{\infty}e^{-(s-a)t}f(t)\,dt.$$The integral on the right is precisely $F(s-a)$, the original transform evaluated at $s-a$ instead of $s$. (A1) This is valid for $s-a>s_0$, i.e. $s>s_0+a$. (R1)右侧积分恰好是 $F(s-a)$，即原变换在 $s-a$ 处的值。(A1) 此结论在 $s-a>s_0$，即 $s>s_0+a$ 时成立。(R1)
By definition, since $u_{c}(t)=0$ for $t
Substitute $\tau=t-c$, so $t=\tau+c$ and $dt=d\tau$; when $t=c$, $\tau=0$. (M1)令 $\tau=t-c$，则 $t=\tau+c$，$dt=d\tau$；当 $t=c$ 时 $\tau=0$。(M1)
$$=\int_{0}^{\infty}e^{-s(\tau+c)}f(\tau)\,d\tau=e^{-cs}\int_{0}^{\infty}e^{-s\tau}f(\tau)\,d\tau=e^{-cs}F(s).\quad\text{(A1)}$$The change of variable is justified because $f(\tau)$ for $\tau\ge 0$ is the same function; the lower limit becomes $0$ because the original integrand is zero for $t
Apply the first shift theorem (a) with $f(t)=\cos 2t$ and $a=-1$. (M1) The base transform is $F(s)=\mathcal{L}\{\cos 2t\}=\dfrac{s}{s^{2}+4}$. (A1) Replace $s$ by $s-(-1)=s+1$:应用 (a) 的第一平移定理，取 $f(t)=\cos 2t$，$a=-1$。(M1) 基本变换为 $F(s)=\mathcal{L}\{\cos 2t\}=\dfrac{s}{s^{2}+4}$。(A1) 将 $s$ 替换为 $s-(-1)=s+1$：
$$\mathcal{L}\{e^{-t}\cos 2t\}=F(s+1)=\frac{s+1}{(s+1)^{2}+4}.\quad\text{(A1)}$$Verify the convolution theorem for $1*e^{t}$ (a), and use it to invert $\tfrac{1}{s(s^{2}+4)}$ without partial fractions (b).验证 $1*e^{t}$ 的卷积定理 (a)，并利用卷积定理不借助部分分式对 $\tfrac{1}{s(s^{2}+4)}$ 求逆变换 (b)。
Compute the convolution directly: (M1)直接计算卷积：(M1)
$$(1*e^{t})(t)=\int_{0}^{t}1\cdot e^{\tau}\,d\tau=\Big[e^{\tau}\Big]_{0}^{t}=e^{t}-1.$$(A1) Now take the transform of $e^{t}-1$: $\mathcal{L}\{e^{t}-1\}=\dfrac{1}{s-1}-\dfrac{1}{s}=\dfrac{s-(s-1)}{s(s-1)}=\dfrac{1}{s(s-1)}$. (M1)(A1) 对 $e^{t}-1$ 取变换：$\mathcal{L}\{e^{t}-1\}=\dfrac{1}{s-1}-\dfrac{1}{s}=\dfrac{s-(s-1)}{s(s-1)}=\dfrac{1}{s(s-1)}$。(M1)
The convolution theorem predicts $\mathcal{L}\{1\}\cdot\mathcal{L}\{e^{t}\}=\dfrac{1}{s}\cdot\dfrac{1}{s-1}=\dfrac{1}{s(s-1)}$. This agrees. (A1)卷积定理预测 $\mathcal{L}\{1\}\cdot\mathcal{L}\{e^{t}\}=\dfrac{1}{s}\cdot\dfrac{1}{s-1}=\dfrac{1}{s(s-1)}$，与直接计算结果一致。(A1)
Write $\dfrac{1}{s(s^{2}+4)}=F(s)\cdot G(s)$ with $F(s)=\dfrac{1}{s}$ and $G(s)=\dfrac{1}{s^{2}+4}$. (M1) The inverse transforms are $f(t)=1$ and $g(t)=\dfrac{1}{2}\sin 2t$. (A1)令 $\dfrac{1}{s(s^{2}+4)}=F(s)\cdot G(s)$，其中 $F(s)=\dfrac{1}{s}$，$G(s)=\dfrac{1}{s^{2}+4}$。(M1) 逆变换分别为 $f(t)=1$ 和 $g(t)=\dfrac{1}{2}\sin 2t$。(A1)
The convolution theorem gives: (M1)卷积定理给出：(M1)
$$(f*g)(t)=\int_{0}^{t}1\cdot\frac{1}{2}\sin 2(t-\tau)\,d\tau=\frac{1}{2}\left[\frac{\cos 2(t-\tau)}{2}\right]_{0}^{t}=\frac{1}{4}\Big[\cos 2(t-\tau)\Big]_{0}^{t}.$$Evaluate the limits: at $\tau=t$, $\cos 0=1$; at $\tau=0$, $\cos 2t$. So the expression equals $\dfrac{1}{4}(1-\cos 2t)$. (A1)代入上下限：$\tau=t$ 时 $\cos 0=1$，$\tau=0$ 时 $\cos 2t$。故表达式等于 $\dfrac{1}{4}(1-\cos 2t)$。(A1)
Solve $y''-3y'+2y=4e^{t}$, $y(0)=1$, $y'(0)=0$, by the Laplace transform method.用拉普拉斯变换法求解 $y''-3y'+2y=4e^{t}$，$y(0)=1$，$y'(0)=0$。
Take the Laplace transform of both sides. Applying the derivative formulas $\mathcal{L}\{y'\}=sY-y(0)$ and $\mathcal{L}\{y''\}=s^{2}Y-sy(0)-y'(0)$ with $y(0)=1$ and $y'(0)=0$: (M1)对两侧取拉普拉斯变换。应用导数公式 $\mathcal{L}\{y'\}=sY-y(0)$ 和 $\mathcal{L}\{y''\}=s^{2}Y-sy(0)-y'(0)$，代入 $y(0)=1$，$y'(0)=0$：(M1)
$$(s^{2}Y-s\cdot 1-0)-3(sY-1)+2Y=\frac{4}{s-1}.$$Collect $Y$ terms: $(s^{2}-3s+2)Y = s-3+\dfrac{4}{s-1}$. (A1) Factor the left coefficient as $(s-1)(s-2)$ and combine the right side over the common denominator $s-1$:整理 $Y$ 项：$(s^{2}-3s+2)Y = s-3+\dfrac{4}{s-1}$。(A1) 将左侧系数因式分解为 $(s-1)(s-2)$，并将右侧通分为公分母 $s-1$：
$$(s-1)(s-2)Y=\frac{(s-3)(s-1)+4}{s-1}=\frac{s^{2}-4s+3+4}{s-1}=\frac{s^{2}-4s+7}{s-1}.$$Therefore $Y(s)=\dfrac{s^{2}-4s+7}{(s-1)^{2}(s-2)}$. (A1)因此 $Y(s)=\dfrac{s^{2}-4s+7}{(s-1)^{2}(s-2)}$。(A1)
Write $\dfrac{s^{2}-4s+7}{(s-1)^{2}(s-2)}=\dfrac{A}{s-2}+\dfrac{B}{s-1}+\dfrac{C}{(s-1)^{2}}$. (M1)令 $\dfrac{s^{2}-4s+7}{(s-1)^{2}(s-2)}=\dfrac{A}{s-2}+\dfrac{B}{s-1}+\dfrac{C}{(s-1)^{2}}$。(M1)
Multiply through by $(s-1)^{2}(s-2)$: $s^{2}-4s+7=A(s-1)^{2}+B(s-1)(s-2)+C(s-2)$.两边乘以 $(s-1)^{2}(s-2)$：$s^{2}-4s+7=A(s-1)^{2}+B(s-1)(s-2)+C(s-2)$。
Set $s=2$: $4-8+7=3=A(1)^{2}$, so $A=3$. Set $s=1$: $1-4+7=4=C(1-2)=-C$, so $C=-4$. Compare coefficients of $s^{2}$: $1=A+B$, so $B=1-3=-2$. (M1) Check: constant term: $A\cdot 1+B\cdot(-1)(-2)+C\cdot(-2)=3+(-2)(2)+(-4)(-2)=3-4+8=7$. Correct. (A1)令 $s=2$：$4-8+7=3=A(1)^{2}$，故 $A=3$。令 $s=1$：$1-4+7=4=C(1-2)=-C$，故 $C=-4$。比较 $s^{2}$ 系数：$1=A+B$，故 $B=1-3=-2$。(M1) 验证：常数项：$A\cdot 1+B\cdot(-1)(-2)+C\cdot(-2)=3+(-2)(2)+(-4)(-2)=3-4+8=7$，正确。(A1)
$$Y(s)=\frac{3}{s-2}-\frac{2}{s-1}-\frac{4}{(s-1)^{2}}.$$Recall $\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}$ and $\mathcal{L}^{-1}\{1/(s-a)^{2}\}=te^{at}$. (M1)利用 $\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}$ 和 $\mathcal{L}^{-1}\{1/(s-a)^{2}\}=te^{at}$。(M1)
$$y(t)=3e^{2t}-2e^{t}-4te^{t}.$$Verify: $y(0)=3-2-0=1$. (A1) Compute $y'(t)=6e^{2t}-2e^{t}-4e^{t}-4te^{t}=6e^{2t}-6e^{t}-4te^{t}$. Then $y'(0)=6-6-0=0$. Both initial conditions satisfied.验证：$y(0)=3-2-0=1$。(A1) 计算 $y'(t)=6e^{2t}-2e^{t}-4e^{t}-4te^{t}=6e^{2t}-6e^{t}-4te^{t}$，则 $y'(0)=6-6-0=0$，两个初始条件均满足。
Solve $y''+4y=g(t)$, $y(0)=0$, $y'(0)=0$, where $g(t)=0$ for $0\le t<\pi$ and $g(t)=1$ for $t\ge\pi$.求解 $y''+4y=g(t)$，$y(0)=0$，$y'(0)=0$，其中 $g(t)$ 在 $0\le t<\pi$ 时为 $0$，$t\ge\pi$ 时为 $1$。
The forcing switches on at $t=\pi$ and stays on, which is exactly the definition of $u_{\pi}(t)$. (M1) So $g(t)=u_{\pi}(t)$ and by the second shift theorem with $f(t)=1$, $F(s)=1/s$:强迫项在 $t=\pi$ 时接入并保持，恰好是 $u_{\pi}(t)$ 的定义。(M1) 故 $g(t)=u_{\pi}(t)$，由第二平移定理，取 $f(t)=1$，$F(s)=1/s$：
$$\mathcal{L}\{u_{\pi}(t)\cdot 1\}=e^{-\pi s}\cdot\frac{1}{s}.\quad\text{(A1)}$$With $y(0)=0$ and $y'(0)=0$, the transform of the ODE is: (M1)由 $y(0)=0$，$y'(0)=0$，对常微分方程取变换得：(M1)
$$(s^{2}+4)Y(s)=\frac{e^{-\pi s}}{s}.$$Solve: $Y(s)=\dfrac{e^{-\pi s}}{s(s^{2}+4)}$. (M1) Factor out $e^{-\pi s}$ and decompose $\dfrac{1}{s(s^{2}+4)}$ by partial fractions: (A1)解得 $Y(s)=\dfrac{e^{-\pi s}}{s(s^{2}+4)}$。(M1) 提取 $e^{-\pi s}$，对 $\dfrac{1}{s(s^{2}+4)}$ 进行部分分式分解：(A1)
$$\frac{1}{s(s^{2}+4)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+4}.$$Multiply through: $1=A(s^{2}+4)+(Bs+C)s$. Set $s=0$: $1=4A$, so $A=\tfrac{1}{4}$. Compare $s^{2}$: $0=A+B$, so $B=-\tfrac{1}{4}$. Compare $s^{1}$: $0=C$. Thus $\dfrac{1}{s(s^{2}+4)}=\dfrac{1/4}{s}-\dfrac{s/4}{s^{2}+4}=\dfrac{1}{4s}-\dfrac{s}{4(s^{2}+4)}$. (A1)两边乘以分母：$1=A(s^{2}+4)+(Bs+C)s$。令 $s=0$：$1=4A$，故 $A=\tfrac{1}{4}$。比较 $s^{2}$：$0=A+B$，故 $B=-\tfrac{1}{4}$。比较 $s^{1}$：$0=C$。因此 $\dfrac{1}{s(s^{2}+4)}=\dfrac{1/4}{s}-\dfrac{s/4}{s^{2}+4}=\dfrac{1}{4s}-\dfrac{s}{4(s^{2}+4)}$。(A1)
Define $H(s)=\dfrac{1}{4s}-\dfrac{s}{4(s^{2}+4)}$, so $Y(s)=e^{-\pi s}H(s)$. (M1) The inverse of $H(s)$ is $h(t)=\dfrac{1}{4}-\dfrac{1}{4}\cos 2t=\dfrac{1}{4}(1-\cos 2t)$. (M1)令 $H(s)=\dfrac{1}{4s}-\dfrac{s}{4(s^{2}+4)}$，则 $Y(s)=e^{-\pi s}H(s)$。(M1) $H(s)$ 的逆变换为 $h(t)=\dfrac{1}{4}-\dfrac{1}{4}\cos 2t=\dfrac{1}{4}(1-\cos 2t)$。(M1)
By the inverse second shift theorem with $c=\pi$: (A1)由第二平移逆定理，取 $c=\pi$：(A1)
$$y(t)=u_{\pi}(t)\cdot h(t-\pi)=u_{\pi}(t)\cdot\frac{1}{4}(1-\cos 2(t-\pi)).$$Piecewise: $y(t)=0$ for $0\le t<\pi$ and $y(t)=\tfrac{1}{4}(1-\cos 2(t-\pi))$ for $t\ge\pi$. Verify $y(0)=0$: the step gates everything to zero. Verify $y'(0)=0$: likewise. (A1)分段表示：$0\le t<\pi$ 时 $y(t)=0$，$t\ge\pi$ 时 $y(t)=\tfrac{1}{4}(1-\cos 2(t-\pi))$。验证 $y(0)=0$：阶跃函数将一切置为零。验证 $y'(0)=0$：同理。(A1)
Solve $y''+2y'+5y=\delta(t-2)$, $y(0)=0$, $y'(0)=0$, using $\mathcal{L}\{\delta(t-a)\}=e^{-as}$.利用 $\mathcal{L}\{\delta(t-a)\}=e^{-as}$ 求解 $y''+2y'+5y=\delta(t-2)$，$y(0)=0$，$y'(0)=0$。
With $y(0)=0$ and $y'(0)=0$, the transforms of the derivative terms simplify: $\mathcal{L}\{y''\}=s^{2}Y$ and $\mathcal{L}\{y'\}=sY$. (M1) The transform of $\delta(t-2)$ is $e^{-2s}$. So the algebraic equation is:由 $y(0)=0$，$y'(0)=0$，导数项的变换化简为：$\mathcal{L}\{y''\}=s^{2}Y$，$\mathcal{L}\{y'\}=sY$。(M1) $\delta(t-2)$ 的变换为 $e^{-2s}$，代数方程为：
$$(s^{2}+2s+5)Y(s)=e^{-2s}.$$Solve: $Y(s)=\dfrac{e^{-2s}}{s^{2}+2s+5}$. (A1·A1)解得 $Y(s)=\dfrac{e^{-2s}}{s^{2}+2s+5}$。(A1·A1)
Complete the square: $s^{2}+2s+5=(s+1)^{2}+4$. (M1) The denominator has no real roots ($\Delta=4-20<0$), confirming underdamped behavior. The relevant table entry, ignoring the $e^{-2s}$ delay factor, is $\dfrac{1}{(s+1)^{2}+4}$: (M1)配方：$s^{2}+2s+5=(s+1)^{2}+4$。(M1) 分母无实根（$\Delta=4-20<0$），证实为欠阻尼行为。忽略 $e^{-2s}$ 延迟因子，相关变换表项为 $\dfrac{1}{(s+1)^{2}+4}$：(M1)
$$\mathcal{L}^{-1}\!\left\{\frac{1}{(s+1)^{2}+4}\right\}=\frac{1}{2}e^{-t}\sin 2t.\quad\text{(A1)}$$(Factor out $\tfrac{1}{2}$ to match the standard form $\mathcal{L}^{-1}\{b/((s+a)^{2}+b^{2})\}=e^{-at}\sin bt$ with $a=1$, $b=2$.)（提出 $\tfrac{1}{2}$ 以匹配标准形式 $\mathcal{L}^{-1}\{b/((s+a)^{2}+b^{2})\}=e^{-at}\sin bt$，其中 $a=1$，$b=2$。）
Write $Y(s)=e^{-2s}\cdot G(s)$ where $G(s)=\dfrac{1}{(s+1)^{2}+4}$ and $g(t)=\tfrac{1}{2}e^{-t}\sin 2t$ from (b). (M1)令 $Y(s)=e^{-2s}\cdot G(s)$，其中 $G(s)=\dfrac{1}{(s+1)^{2}+4}$，$g(t)=\tfrac{1}{2}e^{-t}\sin 2t$ 由 (b) 给出。(M1)
The inverse second shift theorem with $c=2$ gives: (M1)第二平移逆定理，取 $c=2$，给出：(M1)
$$y(t)=u_{2}(t)\,g(t-2)=u_{2}(t)\cdot\frac{1}{2}e^{-(t-2)}\sin 2(t-2).\quad\text{(A1)}$$For $0\le t<2$: $y(t)=0$ (the unit step gates everything to zero). (A1) Verify $y(0)=0$ and $y'(0)=0$: both hold trivially since $y(t)=0$ on $[0,2)$, consistent with the delta forcing only activating the system at $t=2$.当 $0\le t<2$ 时：$y(t)=0$（单位阶跃函数将一切置为零）。(A1) 验证 $y(0)=0$ 和 $y'(0)=0$：两者显然成立，因为 $y(t)$ 在 $[0,2)$ 上恒为零，与 delta 强迫仅在 $t=2$ 时激活系统的物理图象一致。