Differential Equations - Proof of Newton and Bohlin’s Theorem

• Differential equations play a crucial role in various fields of science and engineering.
• One important topic in differential equations is the study of Newton and Bohlin’s Theorem.
• In this lecture, we will discuss the proof of Newton and Bohlin’s Theorem.
• Understanding these theorems will help us solve complex differential equations efficiently.
• Let’s start by understanding the basic concepts.

Basic Concepts

• Newton’s Theorem: If the sum of two functions 𝑦₁(𝑥) and 𝑦₂(𝑥) is a solution to a linear homogeneous differential equation, then 𝑦₁(𝑥) and 𝑦₂(𝑥) are also individual solutions.
• Bohlin’s Theorem: If 𝑦₁(𝑥) and 𝑦₂(𝑥) are two linearly independent solutions of a linear homogeneous differential equation, then the sum of any linear combination of these solutions is also a solution.
• We will now prove these theorems step by step.

Proof: Newton’s Theorem

1. Assume 𝑦₁(𝑥) + 𝑦₂(𝑥) = 𝑦(𝑥) is a solution to the given differential equation.

2. Let 𝑎𝑥²𝑦’’ + 𝑏𝑥𝑦’ + 𝑐𝑦 = 0 be the differential equation.

3. Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation.

4. Simplify the equation using the linearity property of differentiation.

5. Rearrange the terms and substitute 𝑦₁(𝑥) and 𝑦₂(𝑥) separately.

6. Finally, we get the equation 𝑎𝑥²(𝑦₁’’ + 𝑦₂’’) + 𝑏𝑥(𝑦₁’ + 𝑦₂’) + 𝑐(𝑦₁ + 𝑦₂) = 0.

7. Simplify further and observe that 𝑦₁(𝑥) and 𝑦₂(𝑥) satisfy the given linear homogeneous differential equation individually.

8. Hence, the theorem is proved. 𝑒𝑛𝑑.

Proof: Bohlin’s Theorem

1. Assume 𝑦₁(𝑥) and 𝑦₂(𝑥) are linearly independent solutions of the given differential equation.

2. Consider any linear combination of 𝑦₁(𝑥) and 𝑦₂(𝑥) of the form 𝐶₁𝑦₁(𝑥) + 𝐶₂𝑦₂(𝑥), where 𝐶₁ and 𝐶₂ are constants.

3. We need to prove that this combination is also a solution to the differential equation.

4. Substitute the linear combination into the differential equation and simplify.

5. Rewrite the equation using the linearity property of differentiation.

6. Rearrange the terms and distribute the constants 𝐶₁ and 𝐶₂.

7. Observe that each term in the resulting equation is zero since 𝑦₁(𝑥) and 𝑦₂(𝑥) satisfy the given differential equation individually.

8. This implies that the linear combination 𝐶₁𝑦₁(𝑥) + 𝐶₂𝑦₂(𝑥) is also a solution.

9. Hence, the Bohlin’s Theorem is proved. 𝑒𝑛𝑑.

Example 1

• Let’s solve the differential equation 𝑦’’ - 4𝑦’ + 3𝑦 = 0 using Newton’s Theorem.
• Assume 𝑦₁ = 𝑒^𝑥 and 𝑦₂ = 𝑒^3𝑥 are two solutions.
• By Newton’s Theorem, 𝑦 = 𝑦₁ + 𝑦₂ should also be a solution.
• Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation and simplify.
• The resulting equation 𝑒^𝑥 - 2𝑒^3𝑥 + 3𝑒^𝑥 = 0 can be simplified further.
• Using algebraic manipulations, we get (4𝑒^𝑥 - 2)𝑒^2𝑥 = 0.
• This implies that either 𝑦₁ = 𝐶𝑒^𝑥 or 𝑦₂ = 𝐶𝑒^3𝑥 can be solutions.

Example 2

• Let’s solve the differential equation 𝑦’’ - 4𝑦’ + 4𝑦 = 0 using Bohlin’s Theorem.
• Assume 𝑦₁ = 𝑒^2𝑥 and 𝑦₂ = 𝑥𝑒^2𝑥 are two solutions.
• By Bohlin’s Theorem, any linear combination of 𝑦₁ and 𝑦₂ should also be a solution.
• Consider 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, where 𝑐₁ and 𝑐₂ are constants.
• Substitute 𝑦 into the differential equation and simplify.
• The resulting equation (𝑐₂ + 2𝑐₁)𝑒^2𝑥 + (𝑐₂ + 4𝑐₁)𝑥𝑒^2𝑥 = 0 can be simplified further.
• Observing the equation, we can conclude that any value of 𝑐₁ and 𝑐₂ satisfies the differential equation.

These proofs and examples illustrate the concepts and applications of Newton and Bohlin’s Theorems in solving differential equations. Understanding these theorems will help us in solving challenging problems efficiently.

11. Newton’s Theorem - Example 3

• Let’s solve the differential equation 𝑦’’ - 6𝑦’ + 9𝑦 = 0 using Newton’s Theorem.
• Assume 𝑦₁ = 𝑒^3𝑥 and 𝑦₂ = 𝑥𝑒^3𝑥 are two solutions.
• By Newton’s Theorem, 𝑦 = 𝑦₁ + 𝑦₂ should also be a solution.
• Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation and simplify.
• The resulting equation 𝑥(2𝑒^3𝑥 - 3)𝑒^3𝑥 = 0 can be simplified further.

12. Newton’s Theorem - Example 4

• Let’s solve the differential equation 𝑦’’ + 2𝑦’ + 𝑦 = 0 using Newton’s Theorem.
• Assume 𝑦₁ = 𝑒^(-𝑥) and 𝑦₂ = 𝑥𝑒^(-𝑥) are two solutions.
• By Newton’s Theorem, 𝑦 = 𝑦₁ + 𝑦₂ should also be a solution.
• Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation and simplify.
• The resulting equation 2𝑦₁’ + 𝑦₁ + 2𝑦₂’ + 𝑦₂ = 0 can be simplified further.

13. Bohlin’s Theorem - Example 3

• Let’s solve the differential equation 𝑦’’ + 3𝑦’ + 2𝑦 = 0 using Bohlin’s Theorem.
• Assume 𝑦₁ = 𝑒^(-𝑥) and 𝑦₂ = 𝑥𝑒^(-𝑥) are two solutions.
• By Bohlin’s Theorem, any linear combination of 𝑦₁ and 𝑦₂ should also be a solution.
• Consider 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, where 𝑐₁ and 𝑐₂ are constants.
• Substitute 𝑦 into the differential equation and simplify.
• The resulting equation (𝑐₂ + 3𝑐₁)𝑒^(-𝑥) + (2𝑐₂ + 4𝑐₁)𝑥𝑒^(-𝑥) = 0 can be simplified further.

14. Bohlin’s Theorem - Example 4

• Let’s solve the differential equation 𝑦’’ + 3𝑦’ + 2𝑦 = 0 using Bohlin’s Theorem.
• Assume 𝑦₁ = 𝑒^(-𝑥) and 𝑦₂ = 𝑥𝑒^(-𝑥) are two solutions.
• By Bohlin’s Theorem, any linear combination of 𝑦₁ and 𝑦₂ should also be a solution.
• Consider 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, where 𝑐₁ and 𝑐₂ are constants.
• Substitute 𝑦 into the differential equation and simplify.
• The resulting equation (𝑐₂ + 3𝑐₁)𝑒^(-𝑥) + (2𝑐₂ + 4𝑐₁)𝑥𝑒^(-𝑥) = 0 can be simplified further.

15. Newton’s Theorem vs Bohlin’s Theorem

• Newton’s Theorem is applicable when we have one solution and want to find the sum of two solutions.
• Bohlin’s Theorem is applicable when we have two linearly independent solutions and want to find any linear combination of those solutions.
• Newton’s Theorem is derived from Bohlin’s Theorem.
• Bohlin’s Theorem is more general and encompasses Newton’s Theorem.

16. Importance of Newton and Bohlin’s Theorems

• Newton and Bohlin’s Theorems provide valuable tools for solving differential equations.
• By utilizing these theorems, we can find new solutions to differential equations by combining known solutions.
• These theorems also help us understand the linear independence of solutions and their combinations.
• Additionally, Newton and Bohlin’s Theorems have applications in various fields of science and engineering.

17. Summary

• In this lecture, we learned about Newton and Bohlin’s Theorems in differential equations.
• Newton’s Theorem states that the sum of two solutions to a linear homogeneous differential equation is also a solution.
• Bohlin’s Theorem states that any linear combination of linearly independent solutions is also a solution.
• We proved these theorems and solved examples to illustrate their applications.

18. Key Points to Remember

• Newton’s Theorem: The sum of two solutions is also a solution.
• Bohlin’s Theorem: Any linear combination of linearly independent solutions is also a solution.
• Newton’s Theorem is derived from Bohlin’s Theorem.
• These theorems help us find new solutions and understand the linear independence of solutions.

19. Practice Problems

1. Solve 𝑦’’ + 5𝑦’ + 6𝑦 = 0 using Newton’s Theorem.

2. Solve 𝑦’’ - 2𝑦’ + 𝑦 = 0 using Bohlin’s Theorem.

20. Conclusion

• Newton and Bohlin’s Theorems provide powerful tools for solving differential equations.
• Understanding these theorems allows us to find new solutions and analyze the properties of the solutions.
• Make sure to practice solving problems and explore further applications of these theorems.

21. Newton’s Theorem - Example 3

• Let’s solve the differential equation 𝑦’’ + 2𝑦’ + 2𝑦 = 0 using Newton’s Theorem.
• Assume 𝑦₁ = sin 𝑥 and 𝑦₂ = cos 𝑥 are two solutions.
• By Newton’s Theorem, 𝑦 = 𝑦₁ + 𝑦₂ should also be a solution.
• Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation and simplify.
• The resulting equation (1 + sin 𝑥) + (cos 𝑥 - sin 𝑥) = 0 can be simplified further.

22. Newton’s Theorem - Example 4

• Let’s solve the differential equation 𝑦’’ + 2𝑦’ + 𝑦 = 0 using Newton’s Theorem.
• Assume 𝑦₁ = 𝑒^(-𝑥) and 𝑦₂ = 𝑥𝑒^(-𝑥) are two solutions.
• By Newton’s Theorem, 𝑦 = 𝑦₁ + 𝑦₂ should also be a solution.
• Substitute 𝑦 = 𝑦₁ + 𝑦₂ into the differential equation and simplify.
• The resulting equation (𝑥 + 1)𝑒^(-𝑥) = 0 can be simplified further.

23. Bohlin’s Theorem - Example 3

• Let’s solve the differential equation 𝑦’’ - 4𝑦’ + 4𝑦 = 0 using Bohlin’s Theorem.
• Assume 𝑦₁ = 𝑒^2𝑥 and 𝑦₂ = 𝑥𝑒^2𝑥 are two solutions.
• By Bohlin’s Theorem, any linear combination of 𝑦₁ and 𝑦₂ should also be a solution.
• Consider 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, where 𝑐₁ and 𝑐₂ are constants.
• Substitute 𝑦 into the differential equation and simplify.
• The resulting equation (𝑐₂ + 4𝑐₁)𝑒^2𝑥 = 0 can be simplified further.

24. Bohlin’s Theorem - Example 4

• Let’s solve the differential equation 𝑦’’ - 4𝑦’ + 4𝑦 = 0 using Bohlin’s Theorem.
• Assume 𝑦₁ = 𝑒^2𝑥 and 𝑦₂ = 𝑥𝑒^2𝑥 are two solutions.
• By Bohlin’s Theorem, any linear combination of 𝑦₁ and 𝑦₂ should also be a solution.
• Consider 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, where 𝑐₁ and 𝑐₂ are constants.
• Substitute 𝑦 into the differential equation and simplify.
• The resulting equation (𝑐₂ + 4𝑐₁)𝑒^2𝑥 = 0 can be simplified further.

25. Newton’s Theorem vs Bohlin’s Theorem

• Newton’s Theorem is applicable when we have one solution and want to find the sum of two solutions.
• Bohlin’s Theorem is applicable when we have two linearly independent solutions and want to find any linear combination of those solutions.
• Newton’s Theorem is derived from Bohlin’s Theorem.
• Bohlin’s Theorem is more general and encompasses Newton’s Theorem.

26. Importance of Newton and Bohlin’s Theorems

• Newton and Bohlin’s Theorems provide valuable tools for solving differential equations.
• By utilizing these theorems, we can find new solutions to differential equations by combining known solutions.
• These theorems also help us understand the linear independence of solutions and their combinations.
• Additionally, Newton and Bohlin’s Theorems have applications in various fields of science and engineering.

27. Summary

• In this lecture, we learned about Newton and Bohlin’s Theorems in differential equations.
• Newton’s Theorem states that the sum of two solutions to a linear homogeneous differential equation is also a solution.
• Bohlin’s Theorem states that any linear combination of linearly independent solutions is also a solution.
• We proved these theorems and solved examples to illustrate their applications.

28. Key Points to Remember

• Newton’s Theorem: The sum of two solutions is also a solution.
• Bohlin’s Theorem: Any linear combination of linearly independent solutions is also a solution.
• Newton’s Theorem is derived from Bohlin’s Theorem.
• These theorems help us find new solutions and understand the linear independence of solutions.

29. Practice Problems

1. Solve 𝑦’’ + 3𝑦’ + 2𝑦 = 0 using Newton’s Theorem.

2. Solve 𝑦’’ - 5𝑦’ + 6𝑦 = 0 using Bohlin’s Theorem.

30. Conclusion

• Newton and Bohlin’s Theorems provide powerful tools for solving differential equations.
• Understanding these theorems allows us to find new solutions and analyze the properties of the solutions.
• Make sure to practice solving problems and explore further applications of these theorems.