Problem 52 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

The linear combination \(y(x) = c_1 y_1(x) + c_2 y_2(x)\) satisfies the differential equation.

Step 1: Verify solutions for given functions

Start by confirming that both \(y_1(x)\) and \(y_2(x)\) satisfy the linear homogeneous differential equation. Plug \(y_1(x)\) into the given differential equation:\[y_1'' + p(x)y_1' + q(x)y_1 = 0\\] Similarly, substitute \(y_2(x)\):\[y_2'' + p(x)y_2' + q(x)y_2 = 0\\] Since both computations equal zero, \(y_1(x)\) and \(y_2(x)\) are solutions.

Step 2: Define combination function

Define the new function as a linear combination of \(y_1(x)\) and \(y_2(x)\):\[y(x) = c_1 y_1(x) + c_2 y_2(x)\]where \(c_1\) and \(c_2\) are arbitrary constants.

Step 3: Differentiate the combination function

Differentiate \(y(x)\) with respect to \(x\):\[y'(x) = c_1 y_1'(x) + c_2 y_2'(x)\]Then, take 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 \(y''(x)\) back into the original differential equation:\[y'' + p(x)y' + q(x)y = c_1 y_1'' + c_2 y_2'' + p(x)(c_1 y_1' + c_2 y_2') + q(x)(c_1 y_1 + c_2 y_2)\]Simplify each term by distributing \(c_1\) and \(c_2\).

Step 5: Apply principle of superposition

Since both \(y_1(x)\) and \(y_2(x)\) are solutions, substitute their differential equations back in to realize:\[c_1(y_1'' + p(x)y_1' + q(x)y_1) + c_2(y_2'' + p(x)y_2' + q(x)y_2) = 0\]Each expression in parentheses equals zero. Thus, \[y'' + p(x)y' + q(x)y = 0\]implying \(y(x)\) is also a solution.

Key Concepts

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a type of differential equation that can be represented in the form:\[ y'' + p(x)y' + q(x)y = 0 \]In this equation, \(y\) is the dependent variable, \(x\) is the independent variable, and \(p(x)\) and \(q(x)\) are functions of \(x\).
Homogeneous means that every term in the equation involves the dependent variable \(y\) or its derivatives, and there are no standalone terms (i.e., no constant or independent functions on the right side of the equation).
Linear refers to the fact that all instances of \(y\), \(y'\), and \(y''\) appear to the first power and are not multiplied with each other.
This form is foundational in understanding many types of differential equations used in physics and engineering.

Solution Verification

Solution verification is the process of confirming whether a particular function satisfies a given differential equation.
To verify that \(y_1(x)\) and \(y_2(x)\) are solutions of the differential equation, each function is separately substituted into the differential equation.
For instance, when \(y_1(x)\) is substituted:\[ y_1'' + p(x)y_1' + q(x)y_1 = 0 \]If both substituted expressions equate to zero, the function is indeed a solution.
This step ensures that the primary candidates for solutions adhere to the equation's structure, validating their use in further calculations.

Linear Combination of Solutions

The principle of superposition allows any linear combination of solutions for a homogeneous equation to also be a solution.
If \(y_1(x)\) and \(y_2(x)\) are solutions, then:\[ y(x) = c_1 y_1(x) + c_2 y_2(x) \]is also a solution for any constants \(c_1\) and \(c_2\).
This arises because adding or multiplying the solutions by a constant retains the linearity and homogeneity of the equation.
This is a powerful concept as it allows for building more complex solutions from simpler ones, offering flexibility in approach.

Differentiation and Substitution

Differentiation involves finding the derivative of a function, which represents the rate at which the function's value changes with respect to its variable.
For the solution \(y(x) = c_1 y_1(x) + c_2 y_2(x)\), differentiation is applied to obtain the first and second derivatives:

• First derivative: \(y'(x) = c_1 y_1'(x) + c_2 y_2'(x)\)
• Second derivative: \(y''(x) = c_1 y_1''(x) + c_2 y_2''(x)\)

Substitution means plugging these derivatives back into the original differential equation:\[ y'' + p(x)y' + q(x)y \]and ensuring that the equality holds, demonstrating that \(y(x)\) with its derivatives satisfies the original equation.
This formalizes and confirms that the linear combination retains the characteristics needed to be a valid solution.