Problem 43 - A First Course in Differential Equations with Modeling Applications

Question

Consider the differential equation \(a y^{\prime \prime}+b y^{\prime}+c y=e^{k x}\) where \(a, b, c,\) and \(k\) are constants. The auxiliary equation of the associated homogeneous equation is \(a m^{2}+b m+c=0\). (a) If \(k\) is not a root of the auxiliary equation, show that we can find a particular solution of the form \(y_{p}=A e^{k x},\) where \(A=1 /\left(a k^{2}+b k+c\right)\) (b) If \(k\) is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the form \(y_{p}=A x e^{k x},\) where \(A=1 /(2 a k+b) .\) Explain how we know that \(k \neq-b /(2 a)\) (c) If \(k\) is a root of the auxiliary equation of multiplicity two, show that we can find a particular solution of the form \(y=A x^{2} e^{k x},\) where \(A=1 /(2 a)\)

Step-by-Step Solution

Answer

Use the proposed form to find constants: (a) \(A = 1/(ak^2+bk+c)\). (b) \(A = 1/(2ak+b)\) for \(k \, \neq \, -b/(2a)\). (c) \(A = 1/(2a)\).

Step 1: Identifying the Particular Solution Form

To solve the non-homogeneous differential equation \(a y^{\prime \prime}+b y^{\prime}+c y=e^{kx}\), we hypothesize that a particular solution can take the form of \(y_p = A e^{kx}\). The next steps will depend on the relationship between the value of \(k\) and the roots of the auxiliary equation \(a m^2 + b m + c = 0\).

Step 2: Case (a): When \(k\) is Not a Root

If \(k\) is not a root of the auxiliary equation \(a m^2 + b m + c = 0\), we substitute \(y_p = A e^{kx}\) into the differential equation. Differentiate to get \(y_p' = Ak e^{kx}\) and \(y_p'' = Ak^2 e^{kx}\). Substitute these into the lhs of the DE to find \(A\).

Step 3: Solving for A (Case a)

Plugging in the derivatives into \(a y^{\prime\prime} + b y^{\prime} + c y = a(Ak^2e^{kx}) + b(Ake^{kx}) + c(Ae^{kx}) = Ae^{kx}(ak^2 + bk + c)\). To satisfy the equation, this must equal \(e^{kx}\), thus \(A(ak^2 + bk + c) = 1\), giving \(A = \frac{1}{ak^2 + bk + c}\).

Step 4: Case (b): When \(k\) is a Simple Root

If \(k\) is a simple root of multiplicity one of the auxiliary equation, the particular solution must be modified to \(y_p = Ax e^{kx}\). Differentiate to find \(y_p' = Ae^{kx} + Akxe^{kx}\) and \(y_p'' = 2Ake^{kx} + Ak^2xe^{kx}\), and solve similarly.

Step 5: Solving for A (Case b)

Substitute these into the differential equation: \(a(2Ake^{kx} + Ak^2xe^{kx}) + b(Ae^{kx} + Akxe^{kx}) + c(Axe^{kx})\). Collect like terms and solve for \(A\) to satisfy the equation: solve \(2aA + bA = 1\), giving \(A = \frac{1}{2ak + b}\).

Step 6: Explanation for \(k \neq -\frac{b}{2a}\)

If \(k = -\frac{b}{2a}\), the expression \(2ak + b = 0\) becomes undefined for \(A\), meaning \(k\) can't be this value as the solution would not exist.

Step 7: Case (c): When \(k\) is a Root of Multiplicity Two

Here, \(k\) is a double root, necessitating \(y_p = Ax^2e^{kx}\). Differentiate: \(y_p' = 2Axe^{kx} + Akx^2e^{kx}\) and \(y_p'' = 2Ae^{kx} + 2Akxe^{kx} + Ak^2x^2e^{kx}\).

Step 8: Solving for A (Case c)

Substitute derivatives into the differential equation to simplify and meet \(e^{kx}\). Resulting in \(2aA = 1\), which simplifies to \(A = \frac{1}{2a}\).

Key Concepts

Homogeneous Equation

Differential equations can sometimes look a bit daunting. A homogeneous equation is the part of a differential equation without any additional functions or terms. In simpler terms, it's when the equation is set equal to zero. Take for example:

• In our scenario, the homogeneous version of the differential equation is: \[ a y^{\prime \prime}+b y^{\prime}+c y=0 \]

The goal here is to solve this equation to understand the behavior of solutions without external forcing terms. Knowing how to solve this is important, as it allows you to handle the more complicated, non-homogeneous situations by separating them into parts.

When tackling a homogeneous equation, you will often rely upon the auxiliary equation, which we'll discuss next. Understanding both the homogeneous equation and its auxiliary counterpart allows you to tackle the full differential equation with added terms.

Auxiliary Equation

The auxiliary equation stems from the homogeneous differential equation and is key to finding the roots necessary for solving it. In essence, it's a quadratic equation derived from the coefficients of the second, first derivative, and the constant term of the homogeneous differential equation.

Consider:

• The auxiliary equation here is expressed as: \[ a m^{2} + b m + c = 0 \]

This equation is crucial because its solutions, or roots, give us insight into the nature of the solutions of our differential equation. The roots can be real or complex, and these roots significantly influence the form of the solution to both the homogeneous and non-homogeneous equations.

Understanding the roots is especially important when considering whether the external function (like \(e^{kx}\)) appears in this equation. This helps us identify the correct form of the particular solution, which can vary based on these relationships.

Particular Solution

To solve a non-homogeneous differential equation, we don't stop at the homogeneous solution but also find a particular solution that accounts for additional terms in the equation.

This approach gives us:

• Particular solutions take into account the other terms or functions on the other side of the equation, like \(e^{kx}\) in our case.

The relationship between \(k\) and the auxiliary equation's roots determines the form of the particular solution:

• If \(k\) is not a root, we try solutions of the form \(y_{p}=A e^{kx}\).
• If \(k\) is a root of multiplicity one, solutions transform into \(y_{p}=Ax e^{kx}\).
• When \(k\) is a double root, we move to \(y_{p}=Ax^{2} e^{kx}\).

The constants like \(A\) are carefully calculated to ensure this particular solution, combined with any homogeneous solution, satisfies the differential equation fully. Thus, particular solutions play an essential role in describing the full behavior of the system under study.