Answer:
R=[tex]5cos(5t)-3sin(5t)+2[/tex]
Step-by-step explanation:
Using partial fraction decomposition to get a simpler denominator for the inverse transform:
[tex]\frac{7s^2-15s+50}{s(s^2+25)}=\frac{A}{s}+\frac{Bs+C}{s^2+25}[/tex]
Multiple both sides by [tex]s(s^2+25)[/tex]
[tex][s(s^2+25)]\left[\frac{7s^2-15s+50}{s(s^2+25)}\right ]=\left[ \frac{A}{s}+\frac{Bs+C}{s^2+25}\right ][s(s^2+25)]\\7s^2-15s+50=A(s^2+25)+(Bs+C)s\\7s^2-15s+50=As^2+25A+Bs^2+Cs[/tex]
Separate in equations by grouping by the degree of the s:
[tex]7s^2-15s+50=As^2+25A+Bs^2+Cs\\7s^2=As^2+Bs^2\\-15s=Cs\\50=25A[/tex]
Solving for A and C:
[tex]50=25A\\A=\frac{50}{25}=2\\-15s=Cs\\C=\frac{-15s}{s}=-15[/tex]
Substitute the value of A in the first separated equation to find the value of B:
[tex]7s^2=As^2+Bs^2\\7s^2=2s^2+Bs^2\\Bs^2=7s^2-2s^2\\B=\frac{5s^2}{s^2}=5[/tex]
Returning to the partial fraction decomposition:
[tex]\frac{A}{s}+\frac{Bs+C}{s^2+25}=\frac{2}{s}+\frac{5s-15}{s^2+25}[/tex]
Applying the inverse Laplace transform:
[tex]\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=\mathcal{L}^{-1}\left[\frac{2}{s}\right]+\mathcal{L}^{-1} \left[\frac{5s}{s^2+25}\right]+\mathcal{L}^{-1}\left[\frac{-15}{s^2+25}\right]\\[/tex]
[tex]\mathcal{L}^{-1} \left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2\mathcal{L}^{-1}\left[\frac{1}{s}\right]+5\mathcal{L}^{-1} \left[\frac{s}{s^2+5^2}\right]-15\mathcal{L}^{-1}\left[\frac{1}{s^2+5^2}\right][/tex]
Using the next formulas:
[tex]\mathcal{L}^{-1}\left[\frac{1}{s}\right]=1,\mathcal{L}^{-1} \left[\frac{s}{s^2+b^2}\right]=cos(bt),\mathcal{L}^{-1}\left[\frac{1}{s^2+b^2}\right]=\frac{sin(bt)}{b}[/tex]
[tex]\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2(1)+5cos(5t)-15(\frac{sin(5t)}{5} )\\\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2+5cos(5t)-3sin(5t)}\\\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=5cos(5t)-3sin(5t)}+2[/tex]
