Problem 53 - Calculus 3

Question

Prove that if \(a, b,\) and \(c\) are positive constants, then all solutions to the second-order linear differential equation \(a y^{\prime \prime}+b y^{\prime}+c y=0\) approach zero as \(x \rightarrow \infty\) . (Hint: Consider three cases: two distinct roots, repeated real roots, and complex conjugate roots.)

Step-by-Step Solution

Answer

For positive constants, solutions decay to zero for all root cases.

Step 1: Formulate the Characteristic Equation

For the differential equation \( a y'' + b y' + c y = 0 \), the characteristic equation is \( ar^2 + br + c = 0 \). This quadratic equation will help us determine the nature of the roots.

Step 2: Solve the Characteristic Equation

Determine the roots of the characteristic equation using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The nature of these roots will determine the solution behavior of the differential equation.

Step 3: Consider Case 1 - Two Distinct Real Roots

If \(b^2 - 4ac > 0\), the roots are distinct and real: \(r_1\) and \(r_2\). The general solution is \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). For both exponential terms to represent decay, both roots must be negative, which is true given positive constants \(a, b,\) and \(c\).

Step 4: Consider Case 2 - Repeated Real Roots

If \(b^2 - 4ac = 0\), the roots are repeated: \(r_1 = r_2 = -\frac{b}{2a}\). The general solution is \( y(x) = (C_1 + C_2 x) e^{r_1 x} \). The term \( e^{r_1 x} \) ensures decay as long as \( r_1 < 0 \), which holds because \( b > 0 \) and \( a > 0 \).

Step 5: Consider Case 3 - Complex Conjugate Roots

If \(b^2 - 4ac < 0\), the roots are complex conjugates: \( r = \alpha \pm i \beta \), with \( \alpha = -\frac{b}{2a} \). The general solution is \( y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \). Since \( \alpha < 0 \), the envelope \(e^{\alpha x}\) ensures the solution approaches zero as \(x \to \infty\).

Step 6: Conclusion

In all three cases, provided \(a, b,\) and \(c\) are positive, the general solution \(y(x)\) contains exponential decay factors. Therefore, the solution to the differential equation approaches zero as \(x \to \infty\).

Key Concepts

Characteristic Equation

In the realm of second-order linear differential equations, a key step to understanding the solution is formulating the characteristic equation. This is derived from the given differential equation, in our case, written as \( a y'' + b y' + c y = 0 \). The characteristic equation takes the form of a quadratic equation, represented as \( ar^2 + br + c = 0 \). This equation is instrumental in determining the nature of the roots, which in turn dictates the behavior of the differential equation's solutions.

By substituting \(r\) for the derivative term, we bridge the differential equation to an algebraic equation. Solving this quadratic allows us to determine if the solutions involve real or complex roots, which are pivotal in predicting the dynamic response of the solution.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, typically presented as \( ax^2 + bx + c = 0 \). It's the backbone of the characteristic equation discussed earlier. The formula to find the roots of this quadratic equation is \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). These roots are critical because they define the type of solution you'll have for the differential equation.

Depending on the discriminant, \(b^2 - 4ac\), we face three scenarios:

• If \(b^2 - 4ac > 0\), we get two distinct real roots.
• If \(b^2 - 4ac = 0\), we have repeated real roots.
• If \(b^2 - 4ac < 0\), the roots are complex conjugates.

Understanding these roots helps in anticipating the nature of the solutions such as decay, oscillation, or repetitive behaviors.

Complex Conjugate Roots

When dealing with the quadratic equation's roots, an interesting case arises with complex conjugate roots. This occurs if the discriminant \(b^2 - 4ac\) is negative. The roots are then expressed as \( r = \alpha \pm i \beta \), where \( \alpha \) and \( \beta \) are real numbers, and \(i\) is the imaginary unit.

The general solution to the differential equation then transforms into a combination of exponential and trigonometric functions: \[ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \]Here, \(e^{\alpha x}\) acts as an envelope to the trigonometric oscillations. If \(\alpha\) is negative, this envelope decays exponentially, ensuring that the solution approaches zero as \(x\) increases, reflecting exponential decay in response dynamics.

Exponential Decay

Exponential decay is a crucial concept in analyzing solutions to second-order linear differential equations with constant coefficients. When we express solutions using exponentials, such as \(e^{r x}\), the sign of \(r\) dictates whether the function grows or decays.

For the solutions in our differential equation exercise, ensuring exponential decay means verifying that the roots, derived from positive coefficients \(a\), \(b\), and \(c\), lead to a negative exponent term. Negative exponents, imply that the exponential function dwindles to zeros as \(x\) tends to infinity.

For example, in cases with complex conjugate roots where \(\alpha\) is the real part, the term \(e^{\alpha x}\) will govern this behavior. If \(\alpha < 0\), the solution fades away, illustrating why all solutions decline to zero eventually. It's a sign of stability in the system's response, often a desired characteristic in engineering and physics problems.