91Ó°ÊÓ

Second-order linear differential equations involve derivatives up to the second degree. They are crucial in describing many natural phenomena. In general, such equations can be expressed in the form:\[ a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = g(x) \]In our exercise, the equation is simplified as \((D^2 + 9)y = 5e^x - 162x\), where \(D\) represents the differential operator, which simplifies the notation. The designation "second-order" signifies the highest derivative (here, \(D^2\)) in the equation.

• Linear Differential Equation: Terms involving the function and its derivatives appear linearly.
• Highest Order Derivative: The equation involves derivatives up to the second degree (\(D^2\)).

Second-order equations often have solutions involving trigonometric and exponential functions, especially if their characteristic equation yields complex roots.

Differential equations can be categorized based on whether the right-hand side is zero. If a differential equation has the form \(L(y) = 0\), where \(L\) is a linear differential operator, it is homogeneous. Our given equation's homogeneous part is \((D^2 + 9)y = 0\).When the right-hand side of the equation is not zero, as with \((D^2 + 9)y = 5e^x - 162x\), it is non-homogeneous. In such instances, solving the equation involves two main steps:

• Solve the Homogeneous Equation: Find solutions where the equation equals zero.
• Find a Particular Solution: Identify a solution that satisfies the entire non-zero equation.

Combining the homogeneous solution with the particular solution gives the general solution of the non-homogeneous differential equation.

To solve a homogeneous linear differential equation like \((D^2 + 9)y = 0\), we use its characteristic equation. For homogeneous equations of the form \((D^2 + 9)y = 0\), the characteristic equation involves replacing \(D\) with \(r\), leading to \(r^2 + 9 = 0\).

• Finding the Roots: In our example, the roots are \(r = \pm 3i\), which are complex.
• Solution Formulation: Complex roots result in trigonometric solutions involving sine and cosine. Here, the solution became \(y_h = C_1 \cos(3x) + C_2 \sin(3x)\).

The characteristic equation is vital because its roots determine the shape and nature of the solutions to the differential equation.

Finding a particular solution is essential in solving non-homogeneous differential equations. It involves identifying a function that satisfies only the non-homogeneous part of the equation. There are usually one or more terms on the right-hand side which require attention.For the equation \((D^2 + 9)y = 5e^x - 162x\), we need particular solutions for each component. We assumed different forms:

• For \(5e^x\): The term required a particular solution of the form \(Ae^x\).
• For \(-162x\): Here, we assumed a solution of the form \(Bx + C\).

By applying the linear differential operator \((D^2 + 9)\) and equating it to each component, we determined the values of \(A, B,\) and \(C\). Solving these, we found specific values and formulated each particular solution. These were \(\frac{1}{2}e^x\) and \(-18x\). These particular solutions are combined with the homogeneous solution to form the general solution.