Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(\frac{d^{2} r}{d t^{2}}-6 \frac{d r}{d t}+9 r=0\)

Short answer

The differential equation is a second-order linear homogeneous equation with general solution \( r(t) = (C_1 + C_2 t)e^{3t} \).

Step-by-step solution

Identify the Type of Differential Equation

The given equation is \(\frac{d^{2} r}{d t^{2}}-6 \frac{d r}{d t}+9 r=0\). Identify the type of differential equation by examining its form. This is a second-order linear homogeneous differential equation with constant coefficients.

Set Up the Characteristic Equation

For a second-order linear homogeneous differential equation with constant coefficients, the characteristic equation is formed by replacing \( \frac{d r}{d t} \) with \( r' \) and \( \frac{d^{2} r}{d t^{2}} \) with \( r'' \). So, the characteristic equation corresponding to the given differential equation is \( m^2 - 6m + 9 = 0 \).

Solve the Characteristic Equation

Solving the quadratic equation \( m^2 - 6m + 9 = 0 \), we can factor it as \( (m - 3)^2 = 0 \). Therefore, \( m = 3 \) is a repeated root.

Write the General Solution

When the characteristic equation has a repeated root \( m = 3 \), the general solution for the differential equation is of the form \( r(t) = (C_1 + C_2 t) e^{3t} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key concepts

homogeneous differential equation

A homogeneous differential equation is a type of differential equation where all terms are dependent on the function and its derivatives, and there are no external or forcing functions involved. This means that every part of the equation can be written in terms of the function itself and its derivatives, without any constants or other terms added. For example, in the given differential equation \(\frac{d^{2} r}{d t^{2}} - 6 \frac{d r}{d t}+ 9 r = 0\), there are no extra terms added that are independent of the function r(t). An easy way to recognize a homogeneous differential equation is to look for an equal sign with all terms rearranged so they can be expressed with only the function and its derivatives.

characteristic equation

The characteristic equation is a key concept when solving second-order linear homogeneous differential equations with constant coefficients. This equation is derived by assuming a solution of the form \(r(t) = e^{mt}\), where m is a constant. This approach allows us to turn the differential equation into an algebraic equation, which can be solved more easily.
For example, consider the differential equation \(\frac{d^{2} r}{d t^{2}} - 6 \frac{d r}{d t}+ 9 r = 0\). By substituting \(r(t)\) with \(e^{mt}\), we obtain the characteristic equation \((m^2 - 6m + 9 = 0)\). Once the characteristic equation is found, solving for m provides the roots, which are crucial for finding the general solution of the differential equation.

repeated roots solution

When solving the characteristic equation \((m^2 - 6m + 9 = 0)\), we find that we have a repeated root, \(m = 3\). When a differential equation has repeated roots, the general solution form changes slightly to account for this.
The general solution for a second-order differential equation with a repeated root m is given by the formula \(r(t) = (C_1 + C_2 t)e^{mt}\), where \(C_1\) and \(C_2\) are arbitrary constants. This adjustment is necessary because the standard form \(r(t) = C_1 e^{mt} + C_2 e^{nt}\) does not account for repeated roots adequately.
Using our specific example, the root \(m = 3\) results in the general solution \(r(t) = (C_1 + C_2 t)e^{3t}\), effectively capturing all potential solutions to the original differential equation.