Problem 22 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ y^{\prime \prime}-y^{\prime}+11 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = e^{\frac{1}{2}t}(C_1 \cos(\frac{\sqrt{43}}{2} t) + C_2 \sin(\frac{\sqrt{43}}{2} t)) \).

Step 1: Write the Characteristic Equation

The given differential equation is a second-order constant coefficient linear homogeneous differential equation: \( y'' - y' + 11y = 0 \). To find the general solution, assume the solution is of the form \( y = e^{rt} \). Substitute \( y = e^{rt} \), \( y' = re^{rt} \), and \( y'' = r^2e^{rt} \). Then, the characteristic equation is obtained by substituting into the differential equation: \( r^2e^{rt} - re^{rt} + 11e^{rt} = 0 \). Factoring out \( e^{rt} \), the characteristic equation is: \( r^2 - r + 11 = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation \( r^2 - r + 11 = 0 \) is a quadratic equation. To find the roots, use the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = 11 \). Calculate the discriminant: \( b^2 - 4ac = (-1)^2 - 4 \times 1 \times 11 = 1 - 44 = -43 \). The discriminant is negative, which means there are two complex conjugate roots. Calculate the roots: \( r = \frac{1 \pm \sqrt{-43}}{2} = \frac{1 \pm i\sqrt{43}}{2} \).

Step 3: Write the General Solution

Given the complex roots \( r = \frac{1}{2} \pm \frac{i\sqrt{43}}{2} \), the solution to the differential equation can be written in terms of the complex roots. The general solution for complex conjugate roots \( \alpha \pm i\beta \) is \[ y(t) = e^{\alpha t}\left(C_1 \cos(\beta t) + C_2 \sin(\beta t)\right) \]where \( \alpha = \frac{1}{2} \) and \( \beta = \frac{\sqrt{43}}{2} \). Therefore, the general solution is \[ y(t) = e^{\frac{1}{2}t}\left(C_1 \cos\left( \frac{\sqrt{43}}{2} t \right) + C_2 \sin\left( \frac{\sqrt{43}}{2} t \right)\right) \] where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

In the realm of linear differential equations, the characteristic equation is a pivotal concept. It plays an essential role in determining solutions. To derive it, one assumes the solution takes an exponential form, often expressed as \( y = e^{rt} \). For a second-order linear differential equation like the one given, substituting this assumption into the equation leads to terms involving \( r \). The aim is to factor the exponential part, resulting in a polynomial solely in terms of \( r \). This is the characteristic equation. For the example equation, \( y'' - y' + 11y = 0 \), the characteristic equation derived is \( r^2 - r + 11 = 0 \). This polynomial will help find the roots, which are crucial in forming the general solution.

Complex Roots

When tackling the characteristic equation, the type of roots obtained—real or complex—significantly influences the solution. If the discriminant (the expression \( b^2 - 4ac \)) of the polynomial in the characteristic equation is negative, it indicates the presence of complex roots. For instance, in the characteristic equation \( r^2 - r + 11 = 0 \), calculated with a discriminant of \(-43\), two complex roots emerge: \( r = \frac{1}{2} \pm \frac{i\sqrt{43}}{2} \). These roots are complex conjugates because they share a real part and have equal and opposite imaginary parts. Understanding complex roots is vital because they determine the form of the general solution, introducing trigonometric elements.

General Solution

The beauty of solving linear differential equations lies in its general solution, especially with complex roots. For complex roots in the form \( \alpha \pm i\beta \), the solution incorporates exponential and trigonometric functions:

• \( y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \)

Here, \( \alpha \) is the real part and \( \beta \) is related to the imaginary component. In our specific problem, where the roots were \( \frac{1}{2} \pm \frac{i\sqrt{43}}{2} \):

• \( \alpha = \frac{1}{2} \)
• \( \beta = \frac{\sqrt{43}}{2} \)
• The general solution becomes \( y(t) = e^{\frac{1}{2}t}(C_1 \cos(\frac{\sqrt{43}}{2} t) + C_2 \sin(\frac{\sqrt{43}}{2} t)) \).

This format beautifully captures the oscillatory nature introduced by the complex roots.

Quadratic Formula

The quadratic formula is a crucial tool for resolving characteristic equations that emerge from linear differential equations. It provides a direct way to find the roots of any quadratic equation:

• \( ax^2 + bx + c = 0 \)
• Solution given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our linear differential equation problem, after reducing the characteristic equation to \( r^2 - r + 11 = 0 \), the quadratic formula was applied with \( a = 1 \), \( b = -1 \), and \( c = 11 \). The result exhibited a negative discriminant, \( -43 \), leading us to complex roots. The quadratic formula not only finds roots but also reveals their nature—real or complex—thus paving the way to construct a coherent general solution.