Laplace Transform

Laplace Transform — Practice Questions & Solutions

> Perfect for: Semester exams, GATE, ESE, and concept mastery. All solutions are step-by-step and exam-oriented.

Module 1: Definition, Standard Formulas & Existence

Q1. Find the Laplace transform of $f(t) = e^{at}$

Solution:

By definition:

$\mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt$

$\mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-st} \cdot e^{at} \, dt = \int_0^{\infty} e^{-(s-a)t} \, dt$

$= \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_0^{\infty} = 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}$

\boxed{\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \quad s > a}

Q2. Find $\mathcal{L}\{\sin at\}$ and $\mathcal{L}\{\cos at\}$

Solution:

Using Euler's formula: $e^{iat} = \cos at + i\sin at$

$\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia} = \frac{s+ia}{s^2+a^2} = \frac{s}{s^2+a^2} + i\frac{a}{s^2+a^2}$

Equating real and imaginary parts:

$\boxed{\mathcal{L}\{\cos at\} = \frac{s}{s^2+a^2}}$

$\boxed{\mathcal{L}\{\sin at\} = \frac{a}{s^2+a^2}}$

Q3. Find $\mathcal{L}\{t^n\}$ for $n = 0, 1, 2, \dots$

Solution:

$\mathcal{L}\{t^n\} = \int_0^{\infty} e^{-st} t^n \, dt$

Using gamma function: $\int_0^{\infty} e^{-st} t^n \, dt = \frac{\Gamma(n+1)}{s^{n+1}} = \frac{n!}{s^{n+1}}$

\boxed{\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s > 0}

Q4. Find $\mathcal{L}\{\sinh at\}$ and $\mathcal{L}\{\cosh at\}$

Solution:

$\sinh at = \frac{e^{at} - e^{-at}}{2}$

$\mathcal{L}\{\sinh at\} = \frac{1}{2}\left[\frac{1}{s-a} - \frac{1}{s+a}\right] = \frac{1}{2} \cdot \frac{2a}{s^2-a^2}$

$\boxed{\mathcal{L}\{\sinh at\} = \frac{a}{s^2-a^2}}$

$\boxed{\mathcal{L}\{\cosh at\} = \frac{s}{s^2-a^2}}$

Q5. State and verify the existence theorem for $f(t) = e^{3t}$

Solution:

Existence Theorem: If $f(t)$ is piecewise continuous on $[0, \infty)$ and of exponential order $\alpha$ (i.e., $|f(t)| \le Me^{\alpha t}$ for ), then exists for .

Verification for $f(t) = e^{3t}$:

• Piecewise continuous: Yes, $e^{3t}$ is continuous everywhere.
• Exponential order: $|e^{3t}| \le 1 \cdot e^{3t}$, so , .

\mathcal{L}\{e^{3t}\} = \frac{1}{s-3}, \quad s > 3 \quad \checkmark

Module 2: Properties of Laplace Transform

Q6. Find $\mathcal{L}\{e^{2t} \sin 3t\}$

Solution:

First Shifting Theorem: If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$.

Here $f(t) = \sin 3t$, so $F(s) = \frac{3}{s^2+9}$.

With $a = 2$:

$\mathcal{L}\{e^{2t}\sin 3t\} = F(s-2) = \frac{3}{(s-2)^2+9}$

$\boxed{\mathcal{L}\{e^{2t}\sin 3t\} = \frac{3}{(s-2)^2+9}}$

Q7. Find $\mathcal{L}\{t^2 e^{-3t}\}$

Solution:

First, $\mathcal{L}\{t^2\} = \frac{2}{s^3}$.

By first shifting with $a = -3$:

$\mathcal{L}\{t^2 e^{-3t}\} = \frac{2}{(s+3)^3}$

$\boxed{\mathcal{L}\{t^2 e^{-3t}\} = \frac{2}{(s+3)^3}}$

Q8. Find $\mathcal{L}\{t \sin 2t\}$

Solution:

Property: $\mathcal{L}\{t f(t)\} = -\frac{d}{ds}F(s)$

Here $f(t) = \sin 2t$, $F(s) = \frac{2}{s^2+4}$.

$\frac{dF}{ds} = \frac{d}{ds}\left(\frac{2}{s^2+4}\right) = -\frac{4s}{(s^2+4)^2}$

$\mathcal{L}\{t \sin 2t\} = -\frac{dF}{ds} = \frac{4s}{(s^2+4)^2}$

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

Q9. Find $\mathcal{L}\left\{\frac{\sin t}{t}\right\}$

Solution:

Property: $\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_s^{\infty} F(u) \, du$, provided exists.

Here $f(t) = \sin t$, $F(s) = \frac{1}{s^2+1}$.

$\mathcal{L}\left\{\frac{\sin t}{t}\right\} = \int_s^{\infty} \frac{1}{u^2+1} \, du = \left[\tan^{-1} u\right]_s^{\infty}$

$= \frac{\pi}{2} - \tan^{-1} s = \cot^{-1} s$

$\boxed{\mathcal{L}\left\{\frac{\sin t}{t}\right\} = \cot^{-1} s = \tan^{-1}\left(\frac{1}{s}\right)}$

Q10. Find $\mathcal{L}\{\cos^2 t\}$

Solution:

$\cos^2 t = \frac{1 + \cos 2t}{2}$

$\mathcal{L}\{\cos^2 t\} = \frac{1}{2}\mathcal{L}\{1\} + \frac{1}{2}\mathcal{L}\{\cos 2t\}$

$= \frac{1}{2s} + \frac{1}{2} \cdot \frac{s}{s^2+4}$

$= \frac{s^2+4 + s^2}{2s(s^2+4)} = \frac{2s^2+4}{2s(s^2+4)}$

$\boxed{\mathcal{L}\{\cos^2 t\} = \frac{s^2+2}{s(s^2+4)}}$

Q11. Find $\mathcal{L}\{t^3 + 2t^2 - 4t + 5\}$

Solution:

By linearity:

$\mathcal{L}\{t^3\} = \frac{6}{s^4}, \quad \mathcal{L}\{2t^2\} = \frac{4}{s^3}, \quad \mathcal{L}\{-4t\} = -\frac{4}{s^2}, \quad \mathcal{L}\{5\} = \frac{5}{s}$

$\boxed{\mathcal{L}\{t^3 + 2t^2 - 4t + 5\} = \frac{6}{s^4} + \frac{4}{s^3} - \frac{4}{s^2} + \frac{5}{s}}$

Module 3: Laplace Transform of Derivatives & Integrals

Q12. Find $\mathcal{L}\{f'(t)\}$ and $\mathcal{L}\{f''(t)\}$ given $$,

Solution:

Derivative Property:

$\mathcal{L}\{f'(t)\} = sF(s) - f(0) = sF(s) - 2$

$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0) = s^2F(s) - 2s + 1$

$\boxed{\mathcal{L}\{f'(t)\} = sF(s) - 2}$

$\boxed{\mathcal{L}\{f''(t)\} = s^2F(s) - 2s + 1}$

Q13. Find $\mathcal{L}\left\{\int_0^t e^{u} \sin u \, du\right\}$

Solution:

Integral Property: $\mathcal{L}\left\{\int_0^t f(u) \, du\right\} = \frac{F(s)}{s}$

Here $f(t) = e^t \sin t$.

First find $F(s) = \mathcal{L}\{e^t \sin t\}$.

Since $\mathcal{L}\{\sin t\} = \frac{1}{s^2+1}$, by first shifting:

$F(s) = \frac{1}{(s-1)^2+1} = \frac{1}{s^2-2s+2}$

$\mathcal{L}\left\{\int_0^t e^{u}\sin u \, du\right\} = \frac{F(s)}{s} = \frac{1}{s(s^2-2s+2)}$

$\boxed{\mathcal{L}\left\{\int_0^t e^{u}\sin u \, du\right\} = \frac{1}{s(s^2-2s+2)}}$

Module 4: Unit Step Function (Heaviside Function)

Q14. Find $\mathcal{L}\{u(t-a)\}$ where $u(t-a)$ is the unit step function

Solution:

$u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \ge a \end{cases}$

$\mathcal{L}\{u(t-a)\} = \int_0^{\infty} e^{-st} u(t-a) \, dt = \int_a^{\infty} e^{-st} \, dt$

$= \left[\frac{e^{-st}}{-s}\right]_a^{\infty} = \frac{e^{-as}}{s}$

$\boxed{\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}}$

Q15. Find $\mathcal{L}\{f(t)\}$ where $f(t) = \begin{cases} t & 0 \le t < 2 \\ 4-t & 2 \le t < 4 \\ 0 & t \ge 4 \end{cases}$

Solution:

Express using unit step functions:

$f(t) = t[u(t) - u(t-2)] + (4-t)[u(t-2) - u(t-4)]$

$= t \cdot u(t) + (4-t-t)u(t-2) - (4-t)u(t-4)$

$= t \cdot u(t) + (4-2t)u(t-2) + (t-4)u(t-4)$

Rewrite in terms of $(t-2)$ and $(t-4)$:

$4-2t = -2(t-2)$

$t-4 = (t-4)$

So:

$f(t) = t - 2(t-2)u(t-2) + (t-4)u(t-4)$

Now take Laplace:

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

$\mathcal{L}\{(t-2)u(t-2)\} = e^{-2s}\mathcal{L}\{t\} = \frac{e^{-2s}}{s^2}$

$\mathcal{L}\{(t-4)u(t-4)\} = \frac{e^{-4s}}{s^2}$

$\mathcal{L}\{f(t)\} = \frac{1}{s^2} - \frac{2e^{-2s}}{s^2} + \frac{e^{-4s}}{s^2}$

$\boxed{\mathcal{L}\{f(t)\} = \frac{1 - 2e^{-2s} + e^{-4s}}{s^2}}$

Q16. Find $\mathcal{L}\{e^{-2t} u(t-3)\}$

Solution:

We need to express $e^{-2t}$ in terms of $(t-3)$.

$e^{-2t} = e^{-2(t-3+3)} = e^{-6} \cdot e^{-2(t-3)}$

So:

$\mathcal{L}\{e^{-2t}u(t-3)\} = e^{-6} \mathcal{L}\{e^{-2(t-3)}u(t-3)\}$

By second shifting theorem:

$= e^{-6} \cdot e^{-3s} \mathcal{L}\{e^{-2t}\} = e^{-6} \cdot e^{-3s} \cdot \frac{1}{s+2}$

$\boxed{\mathcal{L}\{e^{-2t} u(t-3)\} = \frac{e^{-3(s+2)}}{s+2}}$

Module 5: Laplace Transform of Periodic Functions

Q17. Find the Laplace transform of the square wave with period $2a$:

$f(t) = \begin{cases} 1 & 0 \le t < a \\ -1 & a \le t < 2a \end{cases}$

Solution:

Periodic Function Formula:

$\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-sT}} \int_0^T e^{-st} f(t) \, dt$

Here $T = 2a$.

$\int_0^{2a} e^{-st} f(t) \, dt = \int_0^a e^{-st} \, dt - \int_a^{2a} e^{-st} \, dt$

$= \left[\frac{e^{-st}}{-s}\right]_0^a - \left[\frac{e^{-st}}{-s}\right]_a^{2a}$

$= \frac{1-e^{-as}}{s} - \frac{e^{-as}-e^{-2as}}{s}$

$= \frac{1 - 2e^{-as} + e^{-2as}}{s} = \frac{(1-e^{-as})^2}{s}$

Therefore:

$\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-2as}} \cdot \frac{(1-e^{-as})^2}{s}$

$= \frac{(1-e^{-as})^2}{s(1-e^{-as})(1+e^{-as})} = \frac{1-e^{-as}}{s(1+e^{-as})}$

Multiply numerator and denominator by $e^{as/2}$:

$= \frac{e^{as/2} - e^{-as/2}}{s(e^{as/2} + e^{-as/2})} = \frac{2\sinh(as/2)}{s \cdot 2\cosh(as/2)}$

$\boxed{\mathcal{L}\{f(t)\} = \frac{1}{s}\tanh\left(\frac{as}{2}\right)}$

Q18. Find $\mathcal{L}\{f(t)\}$ where $f(t) = |\sin t|$ (full-wave rectified sine)

Solution:

$|\sin t|$ has period $T = \pi$.

$\mathcal{L}\{|\sin t|\} = \frac{1}{1-e^{-\pi s}} \int_0^{\pi} e^{-st} \sin t \, dt$

Compute the integral:

$\int_0^{\pi} e^{-st}\sin t \, dt = \left[\frac{e^{-st}(-s\sin t - \cos t)}{s^2+1}\right]_0^{\pi}$

$= \frac{e^{-\pi s}(0 - (-1)) - (0-1)}{s^2+1} = \frac{e^{-\pi s} + 1}{s^2+1}$

Therefore:

$\mathcal{L}\{|\sin t|\} = \frac{1+e^{-\pi s}}{(1-e^{-\pi s})(s^2+1)}$

Multiply numerator and denominator by $e^{\pi s/2}$:

$= \frac{e^{\pi s/2} + e^{-\pi s/2}}{(e^{\pi s/2} - e^{-\pi s/2})(s^2+1)} = \frac{2\cosh(\pi s/2)}{2\sinh(\pi s/2)(s^2+1)}$

$\boxed{\mathcal{L}\{|\sin t|\} = \frac{\coth(\pi s/2)}{s^2+1}}$

Module 6: Inverse Laplace Transform

Q19. Find $\mathcal{L}^{-1}\left\{\frac{3s+7}{s^2-2s-3}\right\}$

Solution:

Factor denominator: $s^2-2s-3 = (s-3)(s+1)$

Partial fractions:

$\frac{3s+7}{(s-3)(s+1)} = \frac{A}{s-3} + \frac{B}{s+1}$

$3s+7 = A(s+1) + B(s-3)$

At $s = 3$: $16 = 4A \Rightarrow A = 4$

At $s = -1$: $4 = -4B \Rightarrow B = -1$

$\mathcal{L}^{-1}\left\{\frac{4}{s-3} - \frac{1}{s+1}\right\} = 4e^{3t} - e^{-t}$

$\boxed{\mathcal{L}^{-1}\left\{\frac{3s+7}{s^2-2s-3}\right\} = 4e^{3t} - e^{-t}}$

Q20. Find $\mathcal{L}^{-1}\left\{\frac{s+2}{s^2+4s+8}\right\}$

Solution:

Complete the square:

$s^2+4s+8 = (s+2)^2 + 4 = (s+2)^2 + 2^2$

$\frac{s+2}{(s+2)^2+4}$

By first shifting theorem with $a = -2$:

$\mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} = \cos 2t$

$\mathcal{L}^{-1}\left\{\frac{s+2}{(s+2)^2+4}\right\} = e^{-2t}\cos 2t$

$\boxed{\mathcal{L}^{-1}\left\{\frac{s+2}{s^2+4s+8}\right\} = e^{-2t}\cos 2t}$

Q21. Find $\mathcal{L}^{-1}\left\{\frac{1}{s^2(s+1)}\right\}$

Solution:

Partial fractions:

$\frac{1}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}$

$1 = As(s+1) + B(s+1) + Cs^2$

At $s = 0$: $1 = B$

At $s = -1$: $1 = C$

At $s = 1$: $1 = 2A + 2B + C = 2A + 2 + 1 \Rightarrow A = -1$

$\frac{1}{s^2(s+1)} = -\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1}$

$\mathcal{L}^{-1}\left\{-\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1}\right\} = -1 + t + e^{-t}$

$\boxed{\mathcal{L}^{-1}\left\{\frac{1}{s^2(s+1)}\right\} = t - 1 + e^{-t}}$

Q22. Find $\mathcal{L}^{-1}\left\{\frac{s}{(s^2+4)^2}\right\}$

Solution:

We know $\mathcal{L}\{t \sin 2t\} = \frac{4s}{(s^2+4)^2}$

So:

$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+4)^2}\right\} = \frac{1}{4} t \sin 2t$

Alternatively, using convolution:

$\mathcal{L}^{-1}\left\{\frac{s}{s^2+4} \cdot \frac{1}{s^2+4}\right\} = \cos 2t * \frac{\sin 2t}{2}$

But the multiplication-by-$t$ property is faster here.

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

Module 7: Convolution Theorem

Q23. Using convolution theorem, find $\mathcal{L}^{-1}\left\{\frac{1}{(s+1)(s^2+1)}\right\}$

Solution:

Convolution Theorem: $\mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t) = \int_0^t f(u)g(t-u) \, du$

Let $F(s) = \frac{1}{s+1}$, so $f(t) = e^{-t}$

Let $G(s) = \frac{1}{s^2+1}$, so $g(t) = \sin t$

$f(t) * g(t) = \int_0^t e^{-u} \sin(t-u) \, du$

Using $\sin(t-u) = \sin t \cos u - \cos t \sin u$:

$= \sin t \int_0^t e^{-u}\cos u \, du - \cos t \int_0^t e^{-u}\sin u \, du$

Compute integrals:

$\int e^{-u}\cos u \, du = \frac{e^{-u}(\sin u - \cos u)}{-2}$

$\int e^{-u}\sin u \, du = \frac{e^{-u}(-\sin u - \cos u)}{-2} = \frac{e^{-u}(\sin u + \cos u)}{2}$

Evaluating from $0$ to $t$:

First integral: $\left[\frac{e^{-u}(\sin u - \cos u)}{-2}\right]_0^t = \frac{e^{-t}(\sin t - \cos t)}{-2} - \frac{-1}{-2} = -\frac{e^{-t}(\sin t - \cos t)}{2} - \frac{1}{2}$

Second integral: $\left[\frac{e^{-u}(\sin u + \cos u)}{2}\right]_0^t = \frac{e^{-t}(\sin t + \cos t)}{2} - \frac{1}{2}$

Now:

$= \sin t\left[-\frac{e^{-t}(\sin t - \cos t)}{2} - \frac{1}{2}\right] - \cos t\left[\frac{e^{-t}(\sin t + \cos t)}{2} - \frac{1}{2}\right]$

$= -\frac{e^{-t}\sin t(\sin t - \cos t)}{2} - \frac{\sin t}{2} - \frac{e^{-t}\cos t(\sin t + \cos t)}{2} + \frac{\cos t}{2}$

$= -\frac{e^{-t}}{2}[\sin^2 t - \sin t \cos t + \sin t \cos t + \cos^2 t] + \frac{\cos t - \sin t}{2}$

$= -\frac{e^{-t}}{2}[1] + \frac{\cos t - \sin t}{2}$

$\boxed{\mathcal{L}^{-1}\left\{\frac{1}{(s+1)(s^2+1)}\right\} = \frac{\cos t - \sin t - e^{-t}}{2}}$

Q24. Find $\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+1)}\right\}$ using convolution

Solution:

Let $F(s) = \frac{1}{s^2}$, so $f(t) = t$

Let $G(s) = \frac{1}{s^2+1}$, so $g(t) = \sin t$

$f(t) * g(t) = \int_0^t u \sin(t-u) \, du$

Using integration by parts or expanding:

$= \int_0^t u(\sin t \cos u - \cos t \sin u) \, du$

$= \sin t \int_0^t u \cos u \, du - \cos t \int_0^t u \sin u \, du$

$\int_0^t u \cos u \, du = [u \sin u + \cos u]_0^t = t \sin t + \cos t - 1$

$\int_0^t u \sin u \, du = [-u \cos u + \sin u]_0^t = -t \cos t + \sin t$

$= \sin t(t \sin t + \cos t - 1) - \cos t(-t \cos t + \sin t)$

$= t\sin^2 t + \sin t \cos t - \sin t + t\cos^2 t - \sin t \cos t$

$= t(\sin^2 t + \cos^2 t) - \sin t = t - \sin t$

$\boxed{\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+1)}\right\} = t - \sin t}$

Module 8: Applications to ODEs

Q25. Solve using Laplace transform: $y'' + y = 0$, $y(0) = 1$, $y^{\mathrm{\prime }}($

Solution:

Take Laplace transform:

$s^2Y(s) - sy(0) - y'(0) + Y(s) = 0$

$s^2Y(s) - s + Y(s) = 0$

$(s^2+1)Y(s) = s$

$Y(s) = \frac{s}{s^2+1}$

$\mathcal{L}^{-1}\{Y(s)\} = \cos t$

$\boxed{y(t) = \cos t}$

Q26. Solve: $y'' - 3y' + 2y = e^{3t}$, ,

Solution:

Take Laplace transform:

$s^2Y(s) - s(1) - 0 - 3[sY(s) - 1] + 2Y(s) = \frac{1}{s-3}$

$(s^2 - 3s + 2)Y(s) - s + 3 = \frac{1}{s-3}$

$(s-1)(s-2)Y(s) = \frac{1}{s-3} + s - 3 = \frac{1 + (s-3)^2}{s-3} = \frac{s^2-6s+10}{s-3}$

$Y(s) = \frac{s^2-6s+10}{(s-1)(s-2)(s-3)}$

Partial fractions:

$\frac{s^2-6s+10}{(s-1)(s-2)(s-3)} = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3}$

At $s = 1$: $1-6+10 = 5 = A(1)(-2) \Rightarrow A = -\frac{5}{2}$

At $s = 2$: $4-12+10 = 2 = B(-1)(1) \Rightarrow B = -2$

At $s = 3$: $9-18+10 = 1 = C(2)(1) \Rightarrow C = \frac{1}{2}$

$Y(s) = -\frac{5}{2(s-1)} - \frac{2}{s-2} + \frac{1}{2(s-3)}$

$y(t) = -\frac{5}{2}e^t - 2e^{2t} + \frac{1}{2}e^{3t}$

$\boxed{y(t) = \frac{e^{3t} - 4e^{2t} - 5e^t}{2}}$

Q27. Solve: $y'' + 4y = \sin 2t$, $y(0) = 0$,

Solution:

Laplace transform:

$s^2Y(s) + 4Y(s) = \frac{2}{s^2+4}$

$(s^2+4)Y(s) = \frac{2}{s^2+4}$

$Y(s) = \frac{2}{(s^2+4)^2}$

We know $\mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)^2}\right\} = \frac{\sin at - at\cos at}{2a^3}$

With $a = 2$:

$\mathcal{L}^{-1}\left\{\frac{2}{(s^2+4)^2}\right\} = 2 \cdot \frac{\sin 2t - 2t\cos 2t}{16} = \frac{\sin 2t - 2t\cos 2t}{8}$

$\boxed{y(t) = \frac{\sin 2t - 2t\cos 2t}{8}}$

Q28. Solve: $y'' + 2y' + 5y = e^{-t}\sin t$, ,

Solution:

Laplace transform:

$s^2Y(s) - 1 + 2sY(s) + 5Y(s) = \frac{1}{(s+1)^2+1} = \frac{1}{s^2+2s+2}$

$(s^2+2s+5)Y(s) = 1 + \frac{1}{s^2+2s+2}$

$Y(s) = \frac{1}{(s+1)^2+4} + \frac{1}{[(s+1)^2+4][(s+1)^2+1]}$

Let $u = s+1$:

$Y = \frac{1}{u^2+4} + \frac{1}{(u^2+4)(u^2+1)}$

$= \frac{u^2+1+1}{(u^2+4)(u^2+1)} = \frac{u^2+2}{(u^2+4)(u^2+1)}$

Partial fractions in $u^2$:

$\frac{u^2+2}{(u^2+4)(u^2+1)} = \frac{A}{u^2+4} + \frac{B}{u^2+1}$

$u^2+2 = A(u^2+1) + B(u^2+4)$

At $u^2 = -1$: $1 = 3B \Rightarrow B = \frac{1}{3}$

At $u^2 = -4$: $-2 = -3A \Rightarrow A = \frac{2}{3}$

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

$y(t) = \frac{2}{3}e^{-t}\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} + \frac{1}{3}e^{-t}\mathcal{L}^{-1}\left\{\frac{1}{s^2+1}\right\}$

$\boxed{y(t) = \frac{e^{-t}}{3}\left(\sin t + \frac{2\sin 2t}{2}\right) = \frac{e^{-t}(\sin t + \sin 2t)}{3}}$

Wait, correction: $\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} = \frac{\sin 2t}{2}$

So:

$y(t) = \frac{2}{3}e^{-t} \cdot \frac{\sin 2t}{2} + \frac{1}{3}e^{-t}\sin t$

$\boxed{y(t) = \frac{e^{-t}}{3}(\sin 2t + \sin t)}$

Module 9: Simultaneous Differential Equations

Q29. Solve using Laplace transforms:

$\frac{dx}{dt} + y = 0, \quad \frac{dy}{dt} - x = 0$

with $x(0) = 1$, $y(0) = 0$

Solution:

Take Laplace:

$sX(s) - 1 + Y(s) = 0 \quad \text{...(1)}$

$sY(s) - 0 - X(s) = 0 \quad \text{...(2)}$

From (2): $X(s) = sY(s)$

Substitute into (1):

$s \cdot sY(s) - 1 + Y(s) = 0$

$(s^2+1)Y(s) = 1$