Question

Find the complementary function of each equation: (a) \(y^{\prime \prime}(t)+3 y^{\prime}(t)-4 y=12\) (b) \(y^{\prime \prime}(t)+6 y^{\prime}(t)+5 y=10\) (c) \(y^{\prime \prime}(t)-2 y^{\prime}(t)+y=3\) \((d) y^{\prime \prime}(t)+8 y^{\prime}(t)+16 y=0\)

Short answer

(a) \(y_c(t) = C_1 e^{t} + C_2 e^{-4t}\); (b) \(y_c(t) = C_1 e^{-t} + C_2 e^{-5t}\); (c) \(y_c(t) = (C_1 + C_2 t) e^t\); (d) \(y_c(t) = (C_1 + C_2 t)e^{-4t}\)."

Step-by-step solution

Understanding the Complementary Function

To find the complementary function, we need to solve the associated homogeneous equation for each given differential equation. For a differential equation of the form \(y''(t) + ay'(t) + by = f(t)\), the associated homogeneous equation is \(y''(t) + ay'(t) + by = 0\). We'll solve each part step by step.

Solve (a): Associated Homogeneous Equation

The given equation is \(y''(t) + 3y'(t) - 4y = 12\). The associated homogeneous equation is \(y''(t) + 3y'(t) - 4y = 0\). The characteristic equation is \(r^2 + 3r - 4 = 0\).

Solve (a): Characteristic Equation

To solve the characteristic equation \(r^2 + 3r - 4 = 0\), use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 3\), and \(c = -4\). Calculate the roots: \(r = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\).

Solve (a): Roots and Complementary Function

Calculate the discriminant: \(3^2 + 16 = 25\). The roots are \(r_1 = 1\) and \(r_2 = -4\). Since the roots are real and distinct, the complementary function is: \(y_c(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{t} + C_2 e^{-4t}\).

Solve (b): Associated Homogeneous Equation

Given equation is \(y''(t) + 6y'(t) + 5y = 10\). The associated homogeneous equation is \(y''(t) + 6y'(t) + 5y = 0\). The characteristic equation is \(r^2 + 6r + 5 = 0\).

Solve (b): Characteristic Equation

Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 6\), \(c = 5\). Calculate the roots: \(r = \frac{-6 \pm \sqrt{36 - 20}}{2}\).

Solve (b): Roots and Complementary Function

The discriminant is \(36 - 20 = 16\). The roots are \(r_1 = -1\) and \(r_2 = -5\). Therefore, the complementary function is: \(y_c(t) = C_1 e^{-t} + C_2 e^{-5t}\).

Solve (c): Associated Homogeneous Equation

Given equation is \(y''(t) - 2y'(t) + y = 3\). The associated homogeneous equation is \(y''(t) - 2y'(t) + y = 0\). The characteristic equation is \(r^2 - 2r + 1 = 0\).

Solve (c): Characteristic Equation

The characteristic equation \(r^2 - 2r + 1 = (r - 1)^2 = 0\) gives a repeated root \(r_1 = r_2 = 1\).

Solve (c): Complementary Function

Since there is a repeated root, the complementary function is \(y_c(t) = (C_1 + C_2 t) e^t\).

Solve (d): Associated Homogeneous Equation

Given equation is \(y''(t) + 8y'(t) + 16y = 0\). The characteristic equation is \(r^2 + 8r + 16 = 0\).

Solve (d): Characteristic Equation

The characteristic equation \(r^2 + 8r + 16 = 0\) can be factored as \((r + 4)^2 = 0\), resulting in a repeated root \(r_1 = r_2 = -4\).

Solve (d): Complementary Function

With a repeated root, the complementary function is \(y_c(t) = (C_1 + C_2 t)e^{-4t}\).

Key concepts

Complementary Function

In the world of differential equations, the complementary function is an essential piece of the puzzle that helps us solve linear differential equations, specifically when the solutions describe the natural behavior of the system. When you are given a linear differential equation with constant coefficients, such as \(y''(t) + ay'(t) + by = f(t)\), the complementary function \(y_c(t)\) represents the solution to the associated homogeneous equation \(y''(t) + ay'(t) + by = 0\). The complementary function is typically a combination of exponential functions.

The form of the complementary function depends on the roots of the characteristic equation derived from the associated homogeneous equation. In fact, the nature of these roots dictates the form of \(y_c(t)\):

• If the roots are real and distinct, \(y_c(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\)
• If the roots are repeated, \(y_c(t) = (C_1 + C_2 t) e^{r_1 t}\)
• If the roots are complex, \(y_c(t) = e^{pt} (C_1 \cos(qt) + C_2 \sin(qt))\) where \(r = p \pm qi\)

Each form indicates how the solution behaves over time.

Characteristic Equation

The characteristic equation is a crucial step in finding the complementary function for a differential equation. When you have a second-order differential equation, like \(y''(t) + ay'(t) + by = f(t)\), converting it into a characteristic equation involves assuming a solution of the form \(e^{rt}\). Substituting \(e^{rt}\) into the homogeneous version of the differential equation and factoring out \(e^{rt}\), we can derive the characteristic equation, a polynomial equation based on the coefficients of the original equation.

For example, if you have the differential equation \(y''(t) + 3y'(t) - 4y = 0\), the substitute assumption \(e^{rt}\) transforms it into \(r^2 + 3r - 4 = 0\). Solving this polynomial equation gives the roots, which directly influence the form of the complementary function.

Homogeneous Equation

The homogeneous equation comes into play when solving differential equations by providing a simplified form that does not include an external or forcing function (i.e., one where the right-hand side is zero). The general form is \(y''(t) + ay'(t) + by = 0\).

Solving the homogeneous equation is the first step to finding the complementary function because it reflects the system’s natural response without external input. Once you have the associated homogeneous equation, you can derive the characteristic equation, a vital component in determining the complementary function for the given differential equation. In summary, solving the homogeneous equation allows one to understand the inherent dynamics of the system being modeled, independent of external forces.

Roots of Polynomial Equations

Roots of polynomial equations are the key solutions to the characteristic equation associated with a homogeneous differential equation. These roots provide the essential information needed to form the complementary function.

Solving the quadratic characteristic equation, such as \(r^2 + 3r - 4 = 0\), typically involves factoring, completing the square, or using the quadratic formula. The nature of these roots—whether they are real, repeated, or complex—determines the specific form of the complementary function. For instance:

• Real and distinct roots lead to solutions of the form \(C_1 e^{r_1 t} + C_2 e^{r_2 t}\)
• Repeated roots lead to solutions like \((C_1 + C_2 t) e^{r t}\)
• Complex roots give solutions in the form \(e^{pt} (C_1 \cos(qt) + C_2 \sin(qt))\)

Understanding how to derive and interpret these roots is a foundational skill for resolving differential equations.