Question

For each of the following equations, one solution \(u\) is given. Find the other solution by assuming a solution of the form \(y=u v\). \(x^{3} y^{\prime \prime}+x y^{\prime}-y=0 ; u=x\)

Short answer

The second solution is \(y_2 = 1/x\).

Step-by-step solution

Substitute y using the given solution

Assume a solution of the form: \[ y = uv \] where \( u = x \) and \( v \) is a function to be determined.

Find the derivatives

Calculate the first and second derivatives of \( y \): \[ y' = uv' + u'v = x v' + v \] \[ y'' = u v'' + 2u'v' + u''v = x v'' + 2v' \]

Substitute the derivatives into the original equation

Plug \( y \), \( y' \), and \( y'' \) into the given differential equation: \[ x^{3} (x v'' + 2v') + x (x v' + v) - x = 0 \]

Simplify the equation

Distribute and combine like terms: \[ x^4 v'' + 2x^3 v' + x^2 v' + x v - x = 0 \] Combine the terms: \[ x^4 v'' + (2x^3 + x^2) v' + x v - x = 0 \]

Factor out x

Factor \( x \): \[ x (x^3 v'' + (2x^2 + x) v' + v - 1) = 0 \] Since \( x eq 0 \), we have: \[ x^3 v'' + (2x^2 + x) v' + v - 1 = 0 \]

Simplify the new differential equation

We have a new homogeneous linear differential equation: \[ x^3 v'' + (2x^2 + x) v' + v = 1 \]

Solve the new differential equation

The homogeneous part of the equation is still \[ x^3 v'' + (2x^2 + x) v' + v = 0 \] The solution to this homogeneous equation is \( v = C_1 + C_2 \frac{1}{x} \).

Find the second solution

Since \( y = uv \) and \( u = x \), the second solution is \[ y_2 = x \frac{1}{x} = 1 \] The two linearly independent solutions are \( y_1 = x \) and \( y_2 = \frac{1}{x} \).

Key concepts

Homogeneous Linear Differential Equations

A homogeneous linear differential equation is an equation of the form: \[ a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + ... + a_1(x) y' + a_0(x) y = 0 \] This means that the equation involves derivatives of the function, and all terms are proportional to the function or its derivatives. There are no additional constant or non-homogeneous terms on the right side of the equation, which simplifies the process of finding its solutions.
Essential characteristics of these equations:

• Linearity: The function and its derivatives appear to the first power, without being multiplied together.
• Homogeneity: The right-hand side of the equation is zero; all terms involve the function or its derivatives.

Solution by Reduction of Order

When solving a second-order differential equation with one known solution, a powerful technique is using reduction of order. This means finding a second, linearly independent solution given one known solution. Here’s a general approach:

• Assume a solution of the form \( y = uv \), where \(u\) is the known solution and \(v\) is a new function to determine.
• Compute the first and second derivatives of \( y \).
• Substitute these derivatives into the original differential equation to form a new equation involving \(v\) and its derivatives.
• Simplify the new equation, which usually results in a simpler differential equation that can be solved for \(v\).

In our example, substituting in \( y = x v \) transforms the original equation and allows for a simplified solution path.

First and Second Derivatives

Understanding derivatives is critical for solving differential equations. The first derivative \( y' \) represents the rate of change of the function \( y \), while the second derivative \( y'' \) represents the rate of change of the first derivative, providing insight into the curvature or concavity of \( y \).

In the step-by-step solution:

• We first calculated \( y' \text{ as } uv' + u'v \), which, with \(u = x \) and \(u' = 1 \), simplified to \( x v' + v \).
• Next, we calculated the second derivative \( y'' \text{ as } uv'' + 2u'v' + u''v \), which, using the known values, simplified to \( x v'' + 2v' \).

These derivatives were then substituted back into the original equation, leading to the solution. Knowing how to correctly find and use derivatives is an essential skill in tackling differential equations.