91Ó°ÊÓ

A second-order linear differential equation is an equation involving the second derivative of a function. In simpler terms, it involves taking the rate of the rate of change of a function. These equations often have the form \( a(x)y'' + b(x)y' + c(x)y = f(x) \), where:

• \(y\) is the function of \(x\) we want to solve for.
• \(y'\) is the first derivative.
• \(y''\) is the second derivative.
• \(a(x), b(x), c(x)\) are coefficient functions that depend on \(x\).
• \(f(x)\) is a non-homogeneous term, which can be any function of \(x\).

For example, in our problem, the differential equation is given by \( y'' + 4y = -12 \sin 2x \). This simple structure with constant coefficients (like \(4\) in front of \(y\)) is what makes it linear and second-order. Tackling these equations involves breaking them down into smaller, more manageable parts, such as finding solutions for their homogeneous counterparts first.

Non-homogeneous differential equations have a structure where the right-hand side of the equation is not equal to zero. This contrasts with homogeneous equations where the right-hand side is zero. The form \( y'' + 4y = -12 \sin 2x \) is non-homogeneous because of the \(-12 \sin 2x\) term.
Solving these involves two main steps:

• First, solve the corresponding homogeneous equation \( y'' + 4y = 0 \), which gives us the complementary solution \(y_c\).
• Then, find a particular solution \(y_p\) that accounts for the non-homogeneous part \(-12 \sin 2x\).

Finding \(y_p\) can initially seem tricky. It ideally should mimic the form of \(f(x)\) but oftentimes requires correction, especially if terms overlap with the homogeneous equation’s solution. A suitable form is proposed in the problem statement as \(y_p = Ax \sin 2x + Bx \cos 2x\), which contains factors not present directly in \(y_c\) to ensure uniqueness.

The general solution of a non-homogeneous differential equation is built by combining complementary and particular solutions. Understanding these two concepts is key:

• Complementary Solution \( (y_c) \): This solves the homogeneous version \( y'' + 4y = 0 \). The characteristic equation is simple, \( r^2 + 4 = 0 \), leading to complex roots \( r = \pm 2i \). Therefore, the complementary solution is \(y_c = C_1 \cos 2x + C_2 \sin 2x \), with \(C_1\) and \(C_2\) as arbitrary constants.
• Particular Solution \( (y_p) \): This directly addresses the non-homogeneous part \(-12 \sin 2x\). Using the form \(y_p = Ax \sin 2x + Bx \cos 2x\), upon substituting and equating terms, we solve for \(A\) and \(B\). The calculations yield \(A = 0\) and \(B = 3\), simplifying \(y_p\) to \(3x \cos 2x\).

Putting it all together, the general solution to the equation \( y'' + 4y = -12 \sin 2x \) becomes the sum \( y = y_c + y_p = C_1 \cos 2x + C_2 \sin 2x + 3x \cos 2x \). This combines all possible forms of the solution covering both homogeneous and non-homogeneous aspects.