91Ó°ÊÓ

Linear differential equations are equations that involve derivatives of a function and are linear in terms of that function and its derivatives. These equations can be written in a standard form where the highest derivative term is left on one side of the equality, often expressed in the format:

• \( L(y) = a_n(x) \frac{d^n y}{dx^n} + a_{n-1}(x) \frac{d^{n-1} y}{dx^{n-1}} + ... + a_1(x) \frac{dy}{dx} + a_0(x)y = g(x) \)

where \( a_n(x), a_{n-1}(x), ..., a_0(x) \) are functions of \( x \) and \( g(x) \) is a given function.
This linearity implies that the solutions of these equations can be added together along with constants to form new solutions. In this exercise, you deal with a specific linear differential equation where the function \( L(y) = y'' + \frac{1}{4x^2}y \) is equal to zero.
Here, \( y'' \) denotes the second derivative of \( y \) concerning \( x \), and \( \frac{1}{4x^2} \) suggests that this is a nonhomogeneous term, making the equation a bit more challenging to solve without specific techniques, such as the variation of constants.

Variation of constants is a method for finding particular solutions to non-homogeneous linear differential equations. Traditionally, this method begins with a solution to the corresponding homogeneous equation.
In this problem, you have a solution \( \phi(x) = x^{1/2} \). To find another solution \( \psi(x) \), the variation of constants suggests looking for solutions of the form \( \psi(x) = u(x) \phi(x) \), where \( u(x) \) is an unknown function.

• The goal is to substitute \( \psi(x) = u(x)x^{1/2} \) back into the original differential equation, and by rearranging, solve for \( u(x) \).
• Once you find \( u(x) \), you can determine the general form of the second solution.

Ultimately, the variation of constants method is powerful as it doesn't just provide one solution, but rather opens the path to find a family of solutions. This is why, in this example, the second solution \( \psi(x) \) can involve an arbitrary constant \( C \,\), leading to \( \psi(x) = Cx^{1/2} \).

The product rule is an essential tool in calculus for differentiating the product of two functions. If you have two functions, \( f(x) \) and \( g(x) \), the derivative of their product is given by:

• \( (f\cdot g)' = f'g + fg' \)

In this context, when finding \( \psi(x) = u(x)x^{1/2} \), the product rule is applied to compute its derivatives.
Calculating the first derivative \( \psi'(x) \) involves taking the derivative of the product \( u(x) \cdot x^{1/2} \), leading to:
\[ \psi'(x) = u'(x) \cdot x^{1/2} + u(x) \cdot (\frac{1}{2}x^{-1/2}) \]

• Here, \( u'(x) \) is the derivative of \( u(x) \), and \( \frac{1}{2}x^{-1/2} \) is the derivative of \( x^{1/2} \).

For the second derivative, the product rule is again used in conjunction with the derivative obtained from \( \psi'(x) \). Mastering this technique of applying the product rule aids in simplifying complex expressions encountered in differential equations, such as this one. By accurately breaking down each term's contribution, you can methodically solve the equation and verify solutions.