Question

(with steps please) Find the inverse Laplace transform, f(t), of the function: (7s^2 - 15s + 50)/(s(s^2 + 25))

Answer 1

R=
`5cos(5t)-3sin(5t)+2`

Explanation:

Using partial fraction decomposition to get a simpler denominator for the inverse transform:

`(7s^2-15s+50)/(s(s^2+25))=(A)/(s)+(Bs+C)/(s^2+25)`

Multiple both sides by
`s(s^2+25)`

`[s(s^2+25)]\left[(7s^2-15s+50)/(s(s^2+25))\right ]=\left[ (A)/(s)+(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`

Separate in equations by grouping by the degree of the s:

`7s^2-15s+50=As^2+25A+Bs^2+Cs\\7s^2=As^2+Bs^2\\-15s=Cs\\50=25A`

Solving for A and C:

`50=25A\\A=(50)/(25)=2\\-15s=Cs\\C=(-15s)/(s)=-15`

Substitute the value of A in the first separated equation to find the value of B:

`7s^2=As^2+Bs^2\\7s^2=2s^2+Bs^2\\Bs^2=7s^2-2s^2\\B=(5s^2)/(s^2)=5`

Returning to the partial fraction decomposition:

`(A)/(s)+(Bs+C)/(s^2+25)=(2)/(s)+(5s-15)/(s^2+25)`

Applying the inverse Laplace transform:

`\mathcal{L}^(-1)\left[(2)/(s)+(5s-15)/(s^2+25)\right]=\mathcal{L}^(-1)\left[(2)/(s)\right]+\mathcal{L}^(-1) \left[(5s)/(s^2+25)\right]+\mathcal{L}^(-1)\left[(-15)/(s^2+25)\right]\\`

`\mathcal{L}^(-1) \left[(2)/(s)+(5s-15)/(s^2+25)\right]=2\mathcal{L}^(-1)\left[(1)/(s)\right]+5\mathcal{L}^(-1) \left[(s)/(s^2+5^2)\right]-15\mathcal{L}^(-1)\left[(1)/(s^2+5^2)\right]`

Using the next formulas:

`\mathcal{L}^(-1)\left[(1)/(s)\right]=1,\mathcal{L}^(-1)  \left[(s)/(s^2+b^2)\right]=cos(bt),\mathcal{L}^(-1)\left[(1)/(s^2+b^2)\right]=(sin(bt))/(b)`

`\mathcal{L}^(-1)\left[(2)/(s)+(5s-15)/(s^2+25)\right]=2(1)+5cos(5t)-15((sin(5t))/(5) )\\\mathcal{L}^(-1)\left[(2)/(s)+(5s-15)/(s^2+25)\right]=2+5cos(5t)-3sin(5t)}\\\mathcal{L}^(-1)\left[(2)/(s)+(5s-15)/(s^2+25)\right]=5cos(5t)-3sin(5t)}+2`