Question

Find the complete solution of $$ \left(\frac{d y}{d x}\right)^{2}-\frac{y}{x} \frac{d y}{d x}+\frac{A}{x}=0, $$ where \(A\) is a positive constant.

Short answer

Possible solutions are determined by solving either \[\frac{d y}{d x} = \frac{y + \sqrt{y^2 - 4Ax}}{2x}\] or \[\frac{d y}{d x} = \frac{y - \sqrt{y^2 - 4Ax}}{2x}\].

Step-by-step solution

Identify the form of the differential equation

The given equation is \[\left(\frac{d y}{d x}\right)^{2} - \frac{y}{x} \frac{d y}{d x} + \frac{A}{x} = 0\].This can be recognized as a quadratic equation in terms of \(\frac{d y}{d x}\).

Let v be the first derivative

Let \(v = \frac{d y}{d x}\). Substitute \(v\) into the equation to make it\[v^{2} - \frac{y}{x} v + \frac{A}{x} = 0\].

Solve for v using the quadratic formula

The quadratic formula is given by \[v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here we have: \(a = 1, b = -\frac{y}{x}, c = \frac{A}{x}\). Substituting these into the formula we get: \[v = \frac{\frac{y}{x} \pm \sqrt{\left(\frac{y}{x}\right)^2 - 4 \cdot 1 \cdot \frac{A}{x}}}{2}\].

Simplify the expression for v

Simplify the expression to get: \[v = \frac{y/x \pm \sqrt{(y/x)^2 - 4(A/x)}}{2}\].Combining the fractions and simplifying, we obtain: \[v = \frac{y \pm \sqrt{y^2 - 4Ax}}{2x}\].

Re-substitute v back into the original context

Recall that \(v = \frac{d y}{d x}\), so: \[\frac{d y}{d x} = \frac{y \pm \sqrt{y^2 - 4Ax}}{2x}\].This gives us two possible differential equations to solve: \[\frac{d y}{d x} = \frac{y + \sqrt{y^2 - 4Ax}}{2x}\] and \[\frac{d y}{d x} = \frac{y - \sqrt{y^2 - 4Ax}}{2x}\].

Solve the differential equations

Consider the first differential equation: \[\frac{d y}{d x} = \frac{y + \sqrt{y^2 - 4Ax}}{2x}\]. This can be quite complex and might require special functions for its solution. The second differential equation is: \[\frac{d y}{d x} = \frac{y - \sqrt{y^2 - 4Ax}}{2x}\]. Similarly, this might also require special methods or integration techniques to solve. Generally, consider both cases and analyze the nature of the solutions.

Key concepts

Quadratic Equations

A quadratic equation is any equation of the form ewline$$ax^2 + bx + c = 0,$$ where \(a\), \(b\), and \(c\) are constants. ewlineHere we can see that the original differential equation can be rewritten as a quadratic equation by substituting \(\frac{dy}{dx}\) with \(v\). ewlineQuadratic equations can have either two real solutions, one real solution (a repeated root), or two complex solutions. The nature of the solutions is determined by the discriminant ewline$$b^2 - 4ac.$$ ewlineIn our differential equation, recognizing the quadratic form is crucial for simplifying and solving the equation.

First-Order Differential Equations

First-order differential equations involve the first derivative of the unknown function and can often be written in the form ewline$$\frac{dy}{dx} = f(x, y).$$ ewlineIn our exercise, once we substitute \(\frac{dy}{dx}\) with \(v\), the equation becomes easier to manage as a first-order differential equation. ewlineSolving first-order differential equations can involve various methods: ewline

• Separation of variables
• Integrating factors
• Substitutions

ewlineAfter substituting and simplifying, the separation of variables is often a helpful approach to isolate \(y\) and \(x\) and facilitate easier integration.

Quadratic Formula

The quadratic formula provides a way to solve any quadratic equation of the form ewline$$ax^2 + bx + c = 0.$$ ewlineThe formula is given by ewline$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ ewlineIn our differential equation, substituting \(v = \frac{dy}{dx}\) into the quadratic equation gives us ewline$$v = \frac{\frac{y}{x} \pm \sqrt{(\frac{y}{x})^2 - \frac{4A}{x}}}{2}.$$ ewlineSimplifying this expression yields ewline$$v = \frac{y \pm \sqrt{y^2 - 4Ax}}{2x}.$$ ewlineUsing the quadratic formula in differential equations helps transform complex differential problems into more recognizable algebraic forms.

Integration Techniques

Once we've re-expressed our differential equation, the next step is to solve the resulting first-order differential equations: ewline

• Separate the variables if possible
• Integrate both sides with respect to their respective variables

ewlineConsider the equations: ewline$$\frac{dy}{dx} = \frac{y + \sqrt{y^2 - 4Ax}}{2x}$$ and ewline$$\frac{dy}{dx} = \frac{y - \sqrt{y^2 - 4Ax}}{2x}.$$ ewlineTo integrate these, we need to find a way to simplify the right-hand side to a form where integration is straightforward. This might involve partial fractions, substitution, or recognizing a standard integral form. ewlineIntegration is the last major step in finding the solution to the differential equation, and it's an essential technique for solving various mathematical problems.