Step 1

We use the results from table of Laplace transform.

\(a) f(t)=(t+5)^2+t^2e^{5t} = t^2+10t+25+t^2e^{5t}\)

\(\Rightarrow L\left\{f(t)\right\}=L\left\{t^2+10t+25+t^2e^{5t}\right\}=L\left\{t^2\right\}+10L\left\{t\right\}+L\left\{25\right\}+L\left\{t^2e^{5t}\right\}\)

\(F(s)=\frac{2}{s^3}+\frac{10}{s^2}+\frac{25}{s}+\frac{2}{(s-5)^3}\)

\(b) f(t)=2e^{-t}\cos\left(3t+\frac{\pi}{4}\right) \Rightarrow 2e^{-t}\left[\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - \sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\right]\)

\(\Rightarrow 2e^{-t}\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - 2e^{-t}\sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\)

\(\Rightarrow L\left\{f(t)\right\}=2\cos\left(\frac{\pi}{4}\right)L\left\{e^{-t}\cos(3t)\right\}-2\sin\left(\frac{\pi}{4}\right)L\left\{e^{-t}\sin(t)\right\}\)

we take constant term Laplace

\(\Rightarrow F(s)=\frac{\sqrt2s}{(s+1)^2+9} - \frac{2}{(s+1)^2+9}\)

\(c) F\left(\frac{7s^2-37s+64}{s(s^2-8s+16)}\right)=\frac{4}{s}+\frac{3}{(s-4)}+\frac{7}{(s-4)^2}\)

(by partial fraction)

Then \(L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{4}{s}\right\}+L^{-1}\left\{\frac{3}{(s-4)}\right\}+L^{-1}\left\{\frac{7}{(s-4)^2}\right\}\)

\(f(t)=4+3e^{4t}+7te^{4t}\)

\(d) F(s)=e^{-7s}\left(\frac{1}{s}+\frac{s}{(s^2+1)}\right)=\frac{e^{-7s}}{s} + \frac{s e^{-7s}}{(s^2+7)}\)

\(\Rightarrow L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{e^{-7s}}{s}\right\}+L^{-1}\left\{\frac{s e^{-7s}}{(s^2+7)}\right\}\)

\(=u(t-7)+u(t-7)\cos(t-7)\)

All the results teple of Laplace transform

We use the results from table of Laplace transform.

\(a) f(t)=(t+5)^2+t^2e^{5t} = t^2+10t+25+t^2e^{5t}\)

\(\Rightarrow L\left\{f(t)\right\}=L\left\{t^2+10t+25+t^2e^{5t}\right\}=L\left\{t^2\right\}+10L\left\{t\right\}+L\left\{25\right\}+L\left\{t^2e^{5t}\right\}\)

\(F(s)=\frac{2}{s^3}+\frac{10}{s^2}+\frac{25}{s}+\frac{2}{(s-5)^3}\)

\(b) f(t)=2e^{-t}\cos\left(3t+\frac{\pi}{4}\right) \Rightarrow 2e^{-t}\left[\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - \sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\right]\)

\(\Rightarrow 2e^{-t}\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - 2e^{-t}\sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\)

\(\Rightarrow L\left\{f(t)\right\}=2\cos\left(\frac{\pi}{4}\right)L\left\{e^{-t}\cos(3t)\right\}-2\sin\left(\frac{\pi}{4}\right)L\left\{e^{-t}\sin(t)\right\}\)

we take constant term Laplace

\(\Rightarrow F(s)=\frac{\sqrt2s}{(s+1)^2+9} - \frac{2}{(s+1)^2+9}\)

\(c) F\left(\frac{7s^2-37s+64}{s(s^2-8s+16)}\right)=\frac{4}{s}+\frac{3}{(s-4)}+\frac{7}{(s-4)^2}\)

(by partial fraction)

Then \(L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{4}{s}\right\}+L^{-1}\left\{\frac{3}{(s-4)}\right\}+L^{-1}\left\{\frac{7}{(s-4)^2}\right\}\)

\(f(t)=4+3e^{4t}+7te^{4t}\)

\(d) F(s)=e^{-7s}\left(\frac{1}{s}+\frac{s}{(s^2+1)}\right)=\frac{e^{-7s}}{s} + \frac{s e^{-7s}}{(s^2+7)}\)

\(\Rightarrow L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{e^{-7s}}{s}\right\}+L^{-1}\left\{\frac{s e^{-7s}}{(s^2+7)}\right\}\)

\(=u(t-7)+u(t-7)\cos(t-7)\)

All the results teple of Laplace transform