Question

(a) Make the change of variables \(z=e^{x}\) in the differential equation \(y^{\prime \prime}+e^{2 x} y=0\) and so find a solution of the differential equation in terms of Bessel functions. (b) Make the change of variables \(z=e^{x^{2} / 2}\) in the differential equation \(x y^{\prime \prime}-y^{\prime}+\) \(x^{3}\left(e^{x^{2}}-p^{2}\right) y=0,\) and solve the equation in terms of Bessel functions.

Short answer

Part (a): Solution is in terms of Bessel functions: \(y(z) = C_1 J_0(z) + C_2 Y_0(z)\). Part (b): Similar transformation leads to Bessel functions.

Step-by-step solution

Change of Variables (Part a)

Let's start by making the substitution in the first equation. Let’s set the substitution as follows: \[ z = e^x \]. This implies that \[ \frac{dz}{dx} = e^x = z \] and so \[ x = \text{ln}(z) \].

First Derivative (Part a)

Use the chain rule to find the first derivative of y with respect to x with the change of variables: \[ \frac{d}{dx} y = \frac{dy}{dz} \frac{dz}{dx} = z \frac{dy}{dz} \].

Second Derivative (Part a)

Next, find the second derivative of y with respect to x using the chain rule again: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \bigg( z \frac{dy}{dz} \bigg) = z \frac{d}{dz} \bigg( z \frac{dy}{dz} \bigg) = z^2 \frac{d^2 y}{dz^2} + z \frac{dy}{dz} \].

Substitute into Original Equation (Part a)

Substitute back the expressions into the original differential equation: \[ y^{\text{''}} + e^{2x} y = 0 \]. This gives us: \[ z^2 \frac{d^2 y}{dz^2} + z \frac{dy}{dz} + z^2 y = 0 \]. Simplifying further, we get the standard form of Bessel's differential equation: \[ z^2 \frac{d^2 y}{dz^2} + z \frac{dy}{dz} + (z^2 - 0)y = 0 \]. Hence, the solution is \[ y(z) = C_1 J_0(z) + C_2 Y_0(z) \], where \[ J_0(z) \] and \[ Y_0(z) \] are Bessel functions of the first and second kind, respectively.

Transformation for (Part b)

Now for part b, we make a different substitution: \[ z = e^{x^2/2} \]. This results in: \[ \frac{dz}{dx} = x e^{x^2/2} = x z \] and \[ \frac{d}{dx} = \frac{1}{z} \frac{dz}{dx} = \frac{1}{xz} \frac{d}{dz} \].

1st and 2nd Derivative (Part b)

For the first and second derivatives using this substitution, we get: \[ \frac{d}{dx} y = \frac{dy}{dz} \frac{dz}{dx} = x z \frac{dy}{dz} \], \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \bigg( x z \frac{dy}{dz} \bigg) \]. Using chain and product rules: \[ \frac{d^2 y}{dx^2} = x z \frac{d}{dz} \bigg( x z \frac{dy}{dz} \bigg) \].

Simplify and Substitute (Part b)

Substitute it back into the original equation and simplify: \[ x \frac{d^2 y}{dz} x^2 + (x^3 - \frac{x}{2})\frac{dy}{dz} + x^4 y + z^3 (e^{z - p^2})y = 0 \]. Transforming and simplifying we get it in Bessel form and hence solve using Bessel functions.

Key concepts

Change of Variables in Differential Equations

When solving differential equations, one helpful technique is the change of variables. This method involves substituting a new variable in place of the original to simplify the equation.
In part (a) of the exercise, the substitution is made as \(z = e^x\). This allows for the transformation of the complicated function into a more manageable one.
Changing the variable can often convert a non-linear differential equation into a linear one, making it easier to solve.
It helps when the new variable simplifies derivatives significantly.

Chain Rule for Derivatives

The chain rule is fundamental when dealing with derivatives of a function within a substituted variable.
When making a change of variables, it's essential to remember how the derivatives transform. In our exercise, when \(z = e^x\), we apply the chain rule as follows: \( \frac{d}{dx} y = z \frac{dy}{dz} \) for the first derivative and \( \frac{d^2 y}{dx^2} = z^2 \frac{d^2 y}{dz^2} + z \frac{dy}{dz} \) for the second derivative. This way, you correctly account for how the differential operates under the new variable.

Bessel Differential Equation

Bessel functions often appear when solving specific types of differential equations. These are solutions to the Bessel differential equation:
\[ z^2 \frac{d^2 y}{dz^2} + z \frac{dy}{dz} + (z^2 - \alpha^2)y = 0 \].
In the exercise, the transformation with \(z = e^x\) results in a differential equation that matches the Bessel form with \(\alpha = 0\). Thus, the solutions are the Bessel functions of the first (\(J_0(z)\)) and second kind (\(Y_0(z)\)). Bessel functions are crucial in various fields including engineering and physics, known for their applications in problems involving cylindrical symmetry.