Chapter 16: Problem 6
Find the general solution and also the singular solution, if it exists. $$ p^{2}-x p+y=0. $$
Short Answer
Expert verified
General solutions arise for \(x^2 - 4y > 0\); singular solution is \(y = \frac{x^2}{4}\) when \(x^2 - 4y = 0\).
Step by step solution
01
Understand the Problem
We are given the differential equation \(p^2 - x p + y = 0\), where \(p = \frac{dy}{dx}\). We need to find the general solution and see if a singular solution exists.
02
Substitute for \(p\) and Solve
Substitute \(p = \frac{dy}{dx}\) in the equation \(p^2 - xp + y = 0\). This is a quadratic equation in \(p\). The solutions for \(p\) can be found using the quadratic formula: \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -x, \) and \(c = y\).
03
Calculate the Discriminant
The discriminant \(b^2 - 4ac\) of the quadratic equation is \((-x)^2 - 4 \, (1) \, (y) = x^2 - 4y\). This determines the nature and number of solutions.
04
General Solution Condition
The general solution exists when the discriminant \(x^2 - 4y > 0\). In this case, there are two solutions for \(p\), meaning two particular families of solutions. These correspond to functions satisfying the conditions when \(p = \frac{x \pm \sqrt{x^2 - 4y}}{2}\).
05
Check for Singular Solution
A singular solution might exist if the discriminant \(x^2 - 4y = 0\), since it gives a repeated root (singular solution), where \(p = \frac{x}{2}\). Solve \(x^2 - 4y = 0\) to find the singular solution. Rearranging gives \(y = \frac{x^2}{4}\).
06
Write Down the Solutions
The general solutions are obtained from each value of \(p\) derived from the quadratic formula when \(x^2 - 4y > 0\), and the singular solution is given by \(y = \frac{x^2}{4}\) when \(x^2 - 4y = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution
A general solution to a differential equation is a family of solutions that includes all possible specific solutions a differential equation can have. For the given differential equation \( p^2 - x p + y = 0 \), where \( p = \frac{dy}{dx} \), applying the quadratic formula is key to determining the general solution. The quadratic formula is given by:
- \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Singular Solution
A singular solution of a differential equation is a specific solution that cannot be derived from the general solution by specifying particular values of arbitrary constants. To determine if a singular solution exists for the equation \( p^2 - x p + y = 0 \), we examine the discriminant \( x^2 - 4y \).
- If the discriminant is equal to zero \( x^2 - 4y = 0 \), the equation has a double root.
- This root corresponds to a repeated solution, \( p = \frac{x}{2} \).
Quadratic Equation
Understanding the role of a quadratic equation in finding solutions to differential equations is essential. A quadratic equation has the form \( ax^2 + bx + c = 0 \).
- In our problem, for \( p^2 - x p + y = 0 \), it represents a quadratic equation with respect to \( p \), the derivative \( \frac{dy}{dx} \).
- The coefficients are \( a = 1 \), \( b = -x \), and \( c = y \).
- The quadratic equation provides mechanisms for determining possible values of \( p \) through its solutions.
Discriminant
The discriminant is a crucial component derived from the quadratic equation that informs us about the nature of the equation's roots. It is calculated using the expression \( b^2 - 4ac \).
- For our equation \( p^2 - x p + y = 0 \), which is rewritten as a quadratic in \( p \), the discriminant is \( (-x)^2 - 4(1)(y) = x^2 - 4y \).
- A positive discriminant \( (x^2 - 4y > 0) \) signifies two real and distinct solutions.
- A zero discriminant \( (x^2 - 4y = 0) \) implies a repeated solution, indicating a potential singular solution.
- If the discriminant were negative, it would mean the absence of real solutions for \( p \).