Let's solve the given problem step by step: Given: \[ F(s) = \frac{10s^2 - 35s + 100}{s(s^2 + 25)} \] We need to find the inverse Laplace transform \(\mathcal{L}^{-1}\{F(s)\}\). --- ### **Step 1: Decompose into Partial Fractions** Let: \[ F(s) = \frac{10s^2 - 35s + 100}{s(s^2 + 25)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 25} \] Multiply both sides by \(s(s^2 + 25)\): \[ 10s^2 - 35s + 100 = A(s^2 + 25) + (Bs + C)s \] Expand: \[ 10s^2 - 35s + 100 = A s^2 + 25A + Bs^2 + Cs \] \[ = (A + B)s^2 + Cs + 25A \] Match the powers of \(s\): - \(s^2\) coefficients: \(10 = A + B\) - \(s\) coefficients: \(-35 = C\) - Constant: \(100 = 25A\) --- ### **Step 2: Solve for \(A\), \(B\), \(C\)** From the constant term: \[ 100 = 25A \implies A = 4 \] From the \(s\) term: \[ -35 = C \implies C = -35 \] From the \(s^2\) term: \[ 10 = A + B \implies B = 10 - A = 10 - 4 = 6 \] So: \[ A = 4, \quad B = 6, \quad C = -35 \] --- ### **Step 3: Write the Decomposed Function** \[ F(s) = \frac{4}{s} + \frac{6s - 35}{s^2 + 25} \] --- ### **Step 4: Inverse Laplace Transform Each Term** Recall: - \(\mathcal{L}^{-1}\left\{ \frac{1}{s} \right\} = 1\) - \(\mathcal{L}^{-1}\left\{ \frac{s}{s^2 + a^2} \right\} = \cos(at)\) - \(\mathcal{L}^{-1}\left\{ \frac{a}{s^2 + a^2} \right\} = \sin(at)\) Here, \(a = 5\). So, \[ \mathcal{L}^{-1}\left\{ F(s) \right\} = 4 \cdot 1 + 6 \cos(5t) - 35 \cdot \frac{1}{5} \sin(5t) \] \[ = 4 + 6\cos(5t) - 7\sin(5t) \] --- ### **Final Answer** \[ \boxed{ \mathcal{L}^{-1}\left\{ F(s) \right\} = 4 + 6\cos(5t) - 7\sin(5t) } \]

Let's carefully solve the given problem step-by-step, providing detailed calculations, explanations, and the final answer. --- ### **Given:** \[ F(s) = \frac{10s^2 - 35s + 100}{s(s^2 + 25)} \] We need to find the inverse Laplace transform: \[ \mathcal{L}^{-1}\{F(s)\} \] --- ### **Step 1: Recognize the structure** The denominator is factored as: \[ s(s^2 + 25) \] which suggests that partial fraction decomposition into terms involving \(\frac{A}{s}\) and \(\frac{Bs + C}{s^2 + 25}\) will be appropriate. --- ### **Step 2: Set up partial fractions** Assume: \[ F(s) = \frac{A}{s} + \frac{B s + C}{s^2 + 25} \] Multiply both sides by the denominator \(s(s^2 + 25)\): \[ 10s^2 - 35s + 100 = A(s^2 + 25) + (B s + C)s \] --- ### **Step 3: Expand and equate coefficients** Expand the right side: \[ A s^2 + 25A + B s^2 + C s \] Group like terms: \[ (A + B) s^2 + C s + 25A \] Now, equate coefficients from both sides: \[ \boxed{ \begin{cases} \text{Coefficient of } s^2: & 10 = A + B \\ \text{Coefficient of } s: & -35 = C \\ \text{Constant term}: & 100 = 25A \end{cases} } \] --- ### **Step 4: Solve for A, B, C** From the constant term: \[ 100 = 25A \implies A = \frac{100}{25} = 4 \] From the \(s\) coefficient: \[ C = -35 \] From the \(s^2\) coefficient: \[ 10 = A + B \implies B = 10 - A = 10 - 4 = 6 \] ### **Summary of coefficients:** \[ A = 4, \quad B = 6, \quad C = -35 \] --- ### **Step 5: Rewrite \(F(s)\)** \[ F(s) = \frac{4}{s} + \frac{6s - 35}{s^2 + 25} \] --- ### **Step 6: Find inverse Laplace transforms** Recall standard Laplace inverse transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(a t) \] \[ \mathcal{L}^{-1}\left\{\frac{a}{s^2 + a^2}\right\} = \sin(a t) \] In our case: \[ a^2 = 25 \implies a = 5 \] Express \(\frac{6s - 35}{s^2 + 25}\) as a combination of terms involving \(\frac{s}{s^2 + 25}\) and \(\frac{1}{s^2 + 25}\): \[ \frac{6s - 35}{s^2 + 25} = 6 \cdot \frac{s}{s^2 + 25} - 35 \cdot \frac{1}{s^2 + 25} \] Now, the inverse Laplace transforms are: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2 + 25}\right\} = \cos(5t) \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2 + 25}\right\} = \frac{1}{5} \sin(5t) \] --- ### **Step 7: Final inverse Laplace transform** Putting everything together: \[ \mathcal{L}^{-1}\left\{F(s)\right\} = 4 \cdot 1 + 6 \cos(5t) - 35 \cdot \frac{1}{5} \sin(5t) \] Simplify: \[ = 4 + 6 \cos(5t) - 7 \sin(5t) \] --- ### **### **Final Answer:** \[ \boxed{ \mathcal{L}^{-1}\{F(s)\} = 4 + 6 \cos(5t) - 7 \sin(5t) } \] This is the required time-domain function corresponding to the given Laplace function.

The inverse Laplace transform converts a function from the complex frequency domain back to the time domain, often by decomposing the function into simpler fractions whose transforms are standard. Partial fraction decomposition is used to break complex rational functions into sums of simpler terms involving linear and quadratic denominators, facilitating straightforward application of inverse transforms such as those for exponential, sine, and cosine functions. This systematic approach enables the determination of the original time-dependent function from its Laplace domain representation.