Unit 3 Practice / Review Problems — MATH 2410

Solution:

Start from the definition: \[ \mathcal{L}\{2t\} = \int_0^\infty 2t e^{-st} dt = 2\int_0^\infty t e^{-st} dt. \] Use integration by parts: \[ u = t \Rightarrow du = dt \] \[ dv = e^{-st}dt \Rightarrow v = -\frac{1}{s}e^{-st} \] Apply formula: \[ \int u\,dv = uv - \int v\,du \] \[ \int t e^{-st}dt = -\frac{t}{s}e^{-st} + \frac{1}{s}\int e^{-st}dt \] Compute remaining integral: \[ \int e^{-st}dt = -\frac{1}{s}e^{-st} \] Substitute: \[ = -\frac{t}{s}e^{-st} - \frac{1}{s^2}e^{-st} \] Evaluate from \(0\) to \(\infty\): As \(t \to \infty\): \(e^{-st} \to 0\) At \(t=0\): \[ -\frac{1}{s^2} \] Thus: \[ \int_0^\infty t e^{-st}dt = \frac{1}{s^2} \] Multiply by 2: \[ \boxed{\mathcal{L}\{2t\} = \frac{2}{s^2}} \]

Solution:

\[ \mathcal{L}\{e^{2t}\} = \int_0^\infty e^{2t}e^{-st}dt \] Combine exponents: \[ = \int_0^\infty e^{-(s-2)t}dt \] Integrate: \[ = \left[\frac{-1}{s-2}e^{-(s-2)t}\right]_0^\infty \] As \(t \to \infty\): exponential → 0 At \(t=0\): \[ -\frac{1}{s-2} \] Thus: \[ \boxed{\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}} \]

Solution:

Split: \[ \frac{7s+12}{s^2+16} = 7\frac{s}{s^2+16} + 12\frac{1}{s^2+16} \] Use formulas: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\}=\cos(at) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\}=\frac{1}{a}\sin(at) \] Here \(a=4\): \[ \boxed{7\cos(4t) + 3\sin(4t)} \]

Solution:

Factor: \[ (s+1)(s+2) \] Partial fractions: \[ \frac{s-2}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \] \[ s-2 = A(s+2) + B(s+1) \] \[ = (A+B)s + (2A+B) \] Match: \[ A+B=1,\quad 2A+B=-2 \] Solve: \[ A=-3,\quad B=4 \] \[ \boxed{-3e^{-t} + 4e^{-2t}} \]

Solution:

Factor: \[ s(s^2+4) \] Simplify: \[ \frac{s}{s(s^2+4)} = \frac{1}{s^2+4} \] \[ \boxed{\frac{1}{2}\sin(2t)} \]

Solution:

Factor: \[ (s-3)(s-4) \] Partial fractions: \[ \frac{4s+1}{(s-3)(s-4)} \] Solve: \[ A=-13,\quad B=17 \] \[ \boxed{-13e^{3t} + 17e^{4t}} \]

Solution:

\[ F(s)=\frac{7/2}{s-2} - \frac{5/2}{s-3} + \frac{1}{2}\frac{s-1}{(s-1)^2+1} \] \[ \boxed{ y(t)=\frac{7}{2}e^{2t} - \frac{5}{2}e^{3t} + \frac{1}{2}e^t\cos t } \]

Solution:

Final decomposition: \[ F(s)=\frac{10}{9s^2} + \frac{7}{9s} - \frac{10}{9(s^2+9)} - \frac{7s}{9(s^2+9)} \] \[ \boxed{ y(t)=\frac{10}{9}t + \frac{7}{9} - \frac{10}{27}\sin(3t) - \frac{7}{9}\cos(3t) } \]

Solution:

Use the shifting property: \[ \mathcal{L}\{e^{at}f(t)\} = F(s-a) \] First recall: \[ \mathcal{L}\{t\} = \frac{1}{s^2} \] Apply the shift \(a=3\): \[ \mathcal{L}\{t e^{3t}\} = \frac{1}{(s-3)^2} \] \[ \boxed{\frac{1}{(s-3)^2}} \]

Solution:

First split: \[ (t^2+2t)e^t = t^2 e^t + 2t e^t \] Use: \[ \mathcal{L}\{t^2\} = \frac{2}{s^3}, \quad \mathcal{L}\{t\} = \frac{1}{s^2} \] Apply shifting \(a=1\): \[ \mathcal{L}\{t^2 e^t\} = \frac{2}{(s-1)^3} \] \[ \mathcal{L}\{2t e^t\} = 2\cdot\frac{1}{(s-1)^2} \] Combine: \[ \boxed{\frac{2}{(s-1)^3} + \frac{2}{(s-1)^2}} \]

Solution:

Start with the known transform: \[ \mathcal{L}\{\cos(5t)\} = \frac{s}{s^2+25} \] Use shifting with \(a=-2\): \[ \mathcal{L}\{e^{-2t}\cos(5t)\} = \frac{s+2}{(s+2)^2+25} \] Now use the property: \[ \mathcal{L}\{t f(t)\} = -\frac{d}{ds}F(s) \] Let: \[ F(s) = \frac{s+2}{(s+2)^2+25} \] Differentiate using quotient rule: \[ F'(s) = \frac{ - (s+2)\cdot 2(s+2)}{((s+2)^2+25)^2} \] Simplify numerator: \[ = \frac{(s+2)^2+25 - 2(s+2)^2}{((s+2)^2+25)^2} = \frac{25 - (s+2)^2}{((s+2)^2+25)^2} \] Apply the negative: \[ \mathcal{L}\{t e^{-2t}\cos(5t)\} = -F'(s) = \frac{(s+2)^2 - 25}{((s+2)^2+25)^2} \] \[ \boxed{\frac{(s+2)^2 - 25}{\left((s+2)^2+25\right)^2}} \]

Solution:

We want to rewrite the function in terms of \((t-2)\). Let: \[ t-1 = (t-2)+1 \] So: \[ (t-1)U(t-2) = [(t-2)+1]U(t-2) \] Use the shifting property: \[ \mathcal{L}\{f(t-a)U(t-a)\} = e^{-as}F(s) \] Break into parts: \[ = (t-2)U(t-2) + U(t-2) \] Now: \[ \mathcal{L}\{(t-2)U(t-2)\} = e^{-2s}\mathcal{L}\{t\} = e^{-2s}\frac{1}{s^2} \] \[ \mathcal{L}\{U(t-2)\} = \frac{e^{-2s}}{s} \] Combine: \[ \boxed{e^{-2s}\left(\frac{1}{s^2} + \frac{1}{s}\right)} \]

Solution:

Rewrite using a step function: \[ f(t) = t^2 + U(t-3)\cdot(1) \] (At \(t=3\), the function increases by 1.) Take Laplace transform: \[ \mathcal{L}\{t^2\} = \frac{2}{s^3} \] For the step term: \[ \mathcal{L}\{U(t-3)\} = \frac{e^{-3s}}{s} \] Combine: \[ \boxed{\frac{2}{s^3} + \frac{e^{-3s}}{s}} \]

Solution:

Use: \[ \mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s) \] First: \[ \mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4} \] Let: \[ F(s)=\frac{s}{s^2+4} \] Compute derivatives. First derivative: \[ F'(s)=\frac{(s^2+4)(1) - s(2s)}{(s^2+4)^2} = \frac{s^2+4-2s^2}{(s^2+4)^2} = \frac{4-s^2}{(s^2+4)^2} \] Second derivative: \[ F''(s)=\frac{d}{ds}\left(\frac{4-s^2}{(s^2+4)^2}\right) \] Using quotient rule: \[ = \frac{(-2s)(s^2+4)^2 - (4-s^2)\cdot 2(s^2+4)(2s)}{(s^2+4)^4} \] Factor: \[ = \frac{-2s(s^2+4) - 4s(4-s^2)}{(s^2+4)^3} \] Simplify numerator: \[ = -2s^3 -8s -16s +4s^3 = 2s^3 -24s \] Thus: \[ F''(s)=\frac{2s^3 -24s}{(s^2+4)^3} \] Since \( (-1)^2 = 1 \): \[ \boxed{\frac{2s^3 - 24s}{(s^2+4)^3}} \]

Solution:

First use: \[ \mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9} \] Apply shift \(a=2\): \[ F(s)=\frac{3}{(s-2)^2+9} \] Now use: \[ \mathcal{L}\{t^2 f(t)\} = \frac{d^2}{ds^2}F(s) \] Let \(u=s-2\), so: \[ F(s)=\frac{3}{u^2+9} \] First derivative: \[ F'(s)=\frac{-6u}{(u^2+9)^2} \] Second derivative: \[ F''(s)=\frac{-6(u^2+9)^2 + 24u^2(u^2+9)}{(u^2+9)^4} \] Simplify: \[ = \frac{-6(u^2+9) + 24u^2}{(u^2+9)^3} = \frac{18u^2 -54}{(u^2+9)^3} \] Substitute back \(u=s-2\): \[ \boxed{\frac{18(s-2)^2 - 54}{\left((s-2)^2+9\right)^3}} \]

Solution:

Write as a product: \[ \frac{1}{s(s-1)} = \frac{1}{s}\cdot \frac{1}{s-1} \] Identify: \[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \quad \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^t \] Use convolution: \[ (f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau \] So: \[ y(t) = \int_0^t 1 \cdot e^{t-\tau} d\tau = e^t \int_0^t e^{-\tau} d\tau \] Compute the integral: \[ \int_0^t e^{-\tau} d\tau = 1 - e^{-t} \] Thus: \[ y(t)= e^t(1 - e^{-t}) = e^t - 1 \] \[ \boxed{e^t - 1} \]

Solution:

Write: \[ \frac{1}{s^2(s-1)} = \frac{1}{s^2}\cdot \frac{1}{s-1} \] Identify: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t, \quad \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^t \] Convolution: \[ y(t) = \int_0^t \tau \cdot e^{t-\tau} d\tau = e^t \int_0^t \tau e^{-\tau} d\tau \] Use integration by parts: Let: \[ u=\tau,\quad dv=e^{-\tau}d\tau \] \[ du=d\tau,\quad v=-e^{-\tau} \] Then: \[ \int \tau e^{-\tau} d\tau = -\tau e^{-\tau} + \int e^{-\tau} d\tau = -\tau e^{-\tau} - e^{-\tau} \] Evaluate from \(0\) to \(t\): \[ = [-t e^{-t} - e^{-t}] - [0 -1] = -t e^{-t} - e^{-t} +1 \] Multiply by \(e^t\): \[ y(t)= e^t(1 - e^{-t}(t+1)) = e^t - (t+1) \] \[ \boxed{e^t - t - 1} \]

Solution:

Write: \[ \frac{2}{s^2(s^2+1)} = 2\cdot \frac{1}{s^2}\cdot \frac{1}{s^2+1} \] Identify: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\}=t, \quad \mathcal{L}^{-1}\left\{\frac{1}{s^2+1}\right\}=\sin t \] Convolution: \[ y(t)=2\int_0^t \tau \sin(t-\tau)\,d\tau \] Use identity: \[ \sin(t-\tau)=\sin t \cos \tau - \cos t \sin \tau \] So: \[ y(t)=2\left[ \sin t \int_0^t \tau \cos\tau d\tau - \cos t \int_0^t \tau \sin\tau d\tau \right] \] Compute integrals: \[ \int_0^t \tau \cos\tau d\tau = t\sin t + \cos t -1 \] \[ \int_0^t \tau \sin\tau d\tau = -t\cos t + \sin t \] Substitute: \[ y(t)=2\left[\sin t(t\sin t + \cos t -1) - \cos t(-t\cos t + \sin t)\right] \] Simplify: \[ =2\left[t(\sin^2 t + \cos^2 t) - \sin t\right] =2(t - \sin t) \] \[ \boxed{2t - 2\sin t} \]