Differential Equations: Solved Problems on Laplace Transform Linearity

📚 Introduction to Laplace Transform Linearity

The Laplace Transform is a powerful tool for solving linear differential equations. One of its most fundamental properties is linearity, which allows us to handle complex equations more easily. The linearity property states that the Laplace Transform of a linear combination of functions is the same linear combination of their individual Laplace Transforms.

📜 History and Background

The Laplace Transform is named after Pierre-Simon Laplace, who introduced it in his work on probability theory. Its application to differential equations was later developed by Oliver Heaviside. It has since become a cornerstone in engineering and mathematical analysis.

🔑 Key Principles of Linearity

• ➕ Superposition: The Laplace transform of a sum is the sum of the Laplace transforms: $L[f(t) + g(t)] = L[f(t)] + L[g(t)]$.
• 🔢 Homogeneity: The Laplace transform of a constant times a function is the constant times the Laplace transform of the function: $L[cf(t)] = cL[f(t)]$, where $c$ is a constant.
• 💡 Combined Linearity: Combining both properties: $L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]$, where $a$ and $b$ are constants.

➗ Solved Problems on Laplace Transform Linearity

Problem 1:

Find the Laplace Transform of $f(t) = 3t^2 + 5\sin(2t)$

Solution:

• 🔍 Apply Linearity: $L[3t^2 + 5\sin(2t)] = 3L[t^2] + 5L[\sin(2t)]$
• 📝 Find Individual Transforms: $L[t^2] = \frac{2}{s^3}$ and $L[\sin(2t)] = \frac{2}{s^2 + 4}$
• ✅ Combine Results: $3(\frac{2}{s^3}) + 5(\frac{2}{s^2 + 4}) = \frac{6}{s^3} + \frac{10}{s^2 + 4}$

Problem 2:

Find the Laplace Transform of $f(t) = 2e^{-3t} - 4\cos(t)$

Solution:

• ➕ Apply Linearity: $L[2e^{-3t} - 4\cos(t)] = 2L[e^{-3t}] - 4L[\cos(t)]$
• 💡 Find Individual Transforms: $L[e^{-3t}] = \frac{1}{s + 3}$ and $L[\cos(t)] = \frac{s}{s^2 + 1}$
• 📌 Combine Results: $2(\frac{1}{s + 3}) - 4(\frac{s}{s^2 + 1}) = \frac{2}{s + 3} - \frac{4s}{s^2 + 1}$

Problem 3:

Find the Laplace Transform of $f(t) = 7t + e^{2t} - 3\cosh(t)$

Solution:

• 📐 Apply Linearity: $L[7t + e^{2t} - 3\cosh(t)] = 7L[t] + L[e^{2t}] - 3L[\cosh(t)]$
• ✍️ Find Individual Transforms: $L[t] = \frac{1}{s^2}$, $L[e^{2t}] = \frac{1}{s - 2}$, and $L[\cosh(t)] = \frac{s}{s^2 - 1}$
• 🧪 Combine Results: $7(\frac{1}{s^2}) + \frac{1}{s - 2} - 3(\frac{s}{s^2 - 1}) = \frac{7}{s^2} + \frac{1}{s - 2} - \frac{3s}{s^2 - 1}$

Problem 4:

Find the Laplace Transform of $f(t) = 4\sin(3t) - 2t^3 + 6$

Solution:

• 📝 Apply Linearity: $L[4\sin(3t) - 2t^3 + 6] = 4L[\sin(3t)] - 2L[t^3] + 6L[1]$
• 📖 Find Individual Transforms: $L[\sin(3t)] = \frac{3}{s^2 + 9}$, $L[t^3] = \frac{6}{s^4}$, and $L[1] = \frac{1}{s}$
• 📌 Combine Results: $4(\frac{3}{s^2 + 9}) - 2(\frac{6}{s^4}) + 6(\frac{1}{s}) = \frac{12}{s^2 + 9} - \frac{12}{s^4} + \frac{6}{s}$

Problem 5:

Find the Laplace Transform of $f(t) = 5\cos(2t) + 3e^{-t} - t$

Solution:

• ➕ Apply Linearity: $L[5\cos(2t) + 3e^{-t} - t] = 5L[\cos(2t)] + 3L[e^{-t}] - L[t]$
• 💡 Find Individual Transforms: $L[\cos(2t)] = \frac{s}{s^2 + 4}$, $L[e^{-t}] = \frac{1}{s + 1}$, and $L[t] = \frac{1}{s^2}$
• ✅ Combine Results: $5(\frac{s}{s^2 + 4}) + 3(\frac{1}{s + 1}) - \frac{1}{s^2} = \frac{5s}{s^2 + 4} + \frac{3}{s + 1} - \frac{1}{s^2}$

Problem 6:

Find the Laplace Transform of $f(t) = 2\sinh(t) + 4t^2 - 1$

Solution:

• 📐 Apply Linearity: $L[2\sinh(t) + 4t^2 - 1] = 2L[\sinh(t)] + 4L[t^2] - L[1]$
• ✍️ Find Individual Transforms: $L[\sinh(t)] = \frac{1}{s^2 - 1}$, $L[t^2] = \frac{2}{s^3}$, and $L[1] = \frac{1}{s}$
• 📌 Combine Results: $2(\frac{1}{s^2 - 1}) + 4(\frac{2}{s^3}) - \frac{1}{s} = \frac{2}{s^2 - 1} + \frac{8}{s^3} - \frac{1}{s}$

Problem 7:

Find the Laplace Transform of $f(t) = 6e^{4t} - 2\sin(t) + 5t$

Solution:

• 📝 Apply Linearity: $L[6e^{4t} - 2\sin(t) + 5t] = 6L[e^{4t}] - 2L[\sin(t)] + 5L[t]$
• 📖 Find Individual Transforms: $L[e^{4t}] = \frac{1}{s - 4}$, $L[\sin(t)] = \frac{1}{s^2 + 1}$, and $L[t] = \frac{1}{s^2}$
• ✅ Combine Results: $6(\frac{1}{s - 4}) - 2(\frac{1}{s^2 + 1}) + 5(\frac{1}{s^2}) = \frac{6}{s - 4} - \frac{2}{s^2 + 1} + \frac{5}{s^2}$

🏁 Conclusion

Understanding and applying the linearity property of the Laplace Transform simplifies the process of solving differential equations. By breaking down complex functions into simpler components, we can efficiently find their Laplace Transforms and use them to solve practical problems in various fields.