Problem 13 - Calculus: Early Transcendentals

Question

Find the general solution to the differential equation. $$y^{\prime \prime}+6 y^{\prime}+5 y=4$$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1 e^{-5x} + C_2 e^{-x} + \frac{4}{5}\).

Step 1: Write the Differential Equation

The differential equation given is \(y'' + 6y' + 5y = 4\). We need to find the general solution of this equation, which consists of solving the associated homogeneous equation and finding a particular solution.

Step 2: Solve the Homogeneous Equation

Consider the homogeneous equation, \(y'' + 6y' + 5y = 0\). To solve this, we set up the characteristic equation: \(m^2 + 6m + 5 = 0\).

Step 3: Solve the Characteristic Equation

The characteristic equation is \(m^2 + 6m + 5 = 0\). We can factor it to find \((m + 5)(m + 1) = 0\). Thus, the roots are \(m = -5\) and \(m = -1\).

Step 4: Write General Solution of Homogeneous Equation

The solution to the homogeneous differential equation is given by \(y_h = C_1 e^{-5x} + C_2 e^{-x}\), where \(C_1\) and \(C_2\) are constants.

Step 5: Find Particular Solution

For the non-homogeneous term, which is a constant \(4\), we can try a constant solution \(y_p = A\). Differentiating gives \(y_p' = 0\) and \(y_p''=0\). Substitute into \(y'' + 6y' + 5y = 4\):\(0 + 0 + 5A = 4\) which simplifies to \(5A = 4\). Thus, \(A = \frac{4}{5}\).

Step 6: Combine Homogeneous and Particular Solutions

The general solution \(y(x)\) is the sum of the homogeneous and particular solutions. Therefore, \(y(x) = y_h + y_p = C_1 e^{-5x} + C_2 e^{-x} + \frac{4}{5}\).

Step 7: State the General Solution

The general solution to the differential equation \(y'' + 6y' + 5y = 4\) is \[y(x) = C_1 e^{-5x} + C_2 e^{-x} + \frac{4}{5}\] where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Homogeneous Equations

Differential equations are often composed of two parts: homogeneous equations and non-homogeneous equations. A homogeneous equation is one where all terms involve the function or its derivatives; in other words, there are no standalone constants or functions of the independent variable present on one side of the equation. Consider the given exercise: when solving the equation \(y'' + 6y' + 5y = 4\), we first address the homogeneous part, \(y'' + 6y' + 5y = 0\).

Solving this homogeneous equation involves some important steps:

• First, we construct what is known as the characteristic equation, using the coefficients of the derivatives.
• Then, we solve this equation to find the roots, which helps us to determine the form of the general solution.

This method utilizes principles from algebra, specifically characteristic roots, and helps convert a differential equation into a simpler algebraic format. The solutions to this algebraic equation enable us to express the solution to the differential equation using exponential functions.

Characteristic Equation

The characteristic equation is derived from the coefficients of a linear homogeneous differential equation. When faced with an equation like \(y'' + 6y' + 5y = 0\), we transform it into a characteristic equation by assuming a solution of the form \(y = e^{mx}\).

Upon substitution, the derivatives become powers of \(m\). Consequently, the differential equation translates into an algebraic one: \(m^2 + 6m + 5 = 0\). This is the characteristic equation.
The roots of a characteristic equation determine the form of the solution for the homogeneous differential equation:

• Distinct real roots result in exponential solutions \(y_h = C_1e^{m_1x} + C_2e^{m_2x}\).
• Repeated roots would modify the solution to include a polynomial term, like \(y_h = C_1e^{mx} + C_2xe^{mx}\).

For the example in the exercise, the roots \(m = -5\) and \(m = -1\) led to a distinct exponential solution without requiring additional terms. It's this ability to methodically solve using characteristic equations that simplifies what could otherwise be a complex problem.

Particular Solution

Once the homogeneous equation is solved, it's time to find a particular solution to the non-homogeneous part of the original differential equation. In our exercise, this entails addressing the equation \(y'' + 6y' + 5y = 4\). The term 'particular solution' refers to a specific solution that satisfies the entire non-homogeneous equation.

A smart initial approach is to inspect the non-homogeneous term on the right-hand side. Since it’s a constant (4), we trial a constant particular solution. Thus, we assume a constant form for \(y_p\), i.e., \(y_p = A\). This simple assumption makes differentiation straightforward as both derivatives are zero:

• \(y_p' = 0\)
• \(y_p'' = 0\)

Substituting these into the equation yields:\[5A = 4\]Solving gives the particular solution \(A = \frac{4}{5}\).
The particular solution captures specific behavior due to the non-homogeneous part, and when combined with the homogeneous solution, provides the general solution to the differential equation.