Problem 52 - Calculus 3

Question

(Principle of superposition) Prove that if \(y_{1}(x)\) and \(y_{2}(x)\) are solutions to a linear homogeneous differential equation, \(y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0, \quad\) then the function \(y(x)=c_{1} y_{1}(x)+c_{2} y_{2}(x), \quad\) where \(c_{1}\) and \(c_{2}\) are constants, is also a solution.

Step-by-Step Solution

Answer

The solution is \( y(x) = c_1 y_1(x) + c_2 y_2(x) \).

Step 1: Verify Basic Solutions

We are given that \(y_1(x)\) and \(y_2(x)\) are solutions to the differential equation. This means that they satisfy: 1. \(y_1'' + p(x)y_1' + q(x)y_1 = 0\)2. \(y_2'' + p(x)y_2' + q(x)y_2 = 0\) This forms the basis of using the superposition principle.

Step 2: Define the Combined Solution

We propose a new function \(y(x) = c_1 y_1(x) + c_2 y_2(x)\), where \(c_1\) and \(c_2\) are constants. Our goal is to show \(y(x)\) is also a solution to the differential equation \(y'' + p(x)y' + q(x)y = 0\).

Step 3: Differentiate the Combined Solution

Calculate the first derivative: \( y'(x) = c_1 y_1'(x) + c_2 y_2'(x) \) And the second derivative: \( y''(x) = c_1 y_1''(x) + c_2 y_2''(x) \)

Step 4: Substitute into the Differential Equation

Substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the original equation:\[ c_1 y_1''(x) + c_2 y_2''(x) + p(x)(c_1 y_1'(x) + c_2 y_2'(x)) + q(x)(c_1 y_1(x) + c_2 y_2(x)) = 0 \]

Step 5: Apply Linearity

Use the linearity of the differential equation:We can rewrite it as:\[ c_1 (y_1'' + p(x) y_1' + q(x) y_1) + c_2 (y_2'' + p(x) y_2' + q(x) y_2) \]Since both \(y_1(x)\) and \(y_2(x)\) satisfy the differential equation, each term equals zero:\[ c_1 (0) + c_2 (0) = 0 \]

Step 6: Conclusion

Since substituting \(y(x)\) into the differential equation results in zero, \(y(x)=c_1 y_1(x) + c_2 y_2(x)\) is indeed a solution to the differential equation, showing that the function satisfies the principle of superposition.

Key Concepts

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a type of differential equation where each term is either a function of the variable or its derivatives, and there is no term that is independent of the variable, meaning it equals zero. This kind of equation often takes the form:

• \( y'' + p(x)y' + q(x)y = 0 \)

Here, the coefficients \( p(x) \) and \( q(x) \) are functions of \( x \), and can vary with the input value. The 'homogeneous' part indicates that there is no additional non-zero component added to the equation.
The solutions of such equations are crucial because they explain many natural phenomena, for example, oscillations, population dynamics, and heat conduction.
For students, understanding the structure of a homogeneous differential equation simplifies the solving process, as we assume solutions that make the entire expression zero.

Differential Equation Solutions

Finding solutions to differential equations like \( y'' + p(x)y' + q(x)y = 0 \) involves determining the functions that satisfy the equation across all their domain.
When given two solutions, say \( y_1(x) \) and \( y_2(x) \), they meet the criteria set by the differential equation, like:

• \( y_1'' + p(x)y_1' + q(x)y_1 = 0 \)
• \( y_2'' + p(x)y_2' + q(x)y_2 = 0 \)

To verify solutions involves substituting them back into the original equation to see if it holds true. This process is crucial as it ensures that they indeed solve the equation under study.
Moreover, combining these solutions using constants, as shown in the example of \( c_1 y_1(x) + c_2 y_2(x) \), helps in creating a new solution. This is useful in exploring more complex systems or when initial conditions are applied.

Linearity in Differential Equations

Linearity is a fundamental concept in understanding and solving differential equations. It dictates that the equation and its solutions maintain specific properties that allow combination and scaling.
For a differential equation to be linear, each derivative of the function should appear linearly without any powers or products with other derivatives. The coefficients can be any function of the independent variable but must not be a function of the dependent variable.
This property of linearity carries over to solutions, meaning if \( y_1(x) \) and \( y_2(x) \) are solutions, any linear combination \( c_1 y_1(x) + c_2 y_2(x) \) where \( c_1 \) and \( c_2 \) are constants, also results in a solution.
The principle of superposition directly uses this property: it states that the superposition of solutions yields a new solution, facilitating the analysis and solving of more complex problems by utilizing simpler component solutions.