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Find the general solution and also the singular solution, if it exists. $$ p(x p-y+k)+a=0 $$

Short Answer

Expert verified
The general solution is \( y = cx + ck + \frac{a}{c} \). Singular solution may not exist.

Step by step solution

01

Introduce the problem

We need to find the general solution and any singular solution for the given equation: \( p(x p - y + k) + a = 0 \). The term \( p \) represents \( \frac{dy}{dx} \).
02

Substitute \( p = \frac{dy}{dx} \)

Substitute \( p = \frac{dy}{dx} \) in the equation, giving us \( \frac{dy}{dx}(x \frac{dy}{dx} - y + k) + a = 0 \).
03

Rearrange the equation

To simplify the equation, rearrange it to isolated terms: \( p(xp - y + k) = -a \).
04

Assume a particular form of solution

Consider the form \( y = cx + f(c) \). Then \( p = c \) and substitute them into the equation: \( c(xc - (cx + f(c)) + k) + a = 0 \).
05

Solve for \( f(c) \)

This simplifies to \( c(-f(c) + k) + a = 0 \). Therefore, \( f(c) = c k + \frac{a}{c} \).
06

General solution

Hence, the general solution is \( y = cx + ck + \frac{a}{c} \) for some constant \( c \).
07

Find the singular solution

To find singular solutions, differentiate the original equation: \( p\left(p + x\frac{dp}{dx} \right) - \frac{dy}{dx} + k \frac{dp}{dx} = 0 \). Setting \( p = 0 \), if exists, which leads to solve \( y = ck + \frac{a}{c} \).
08

Identify singular solution existence

If the condition where \( p = 0 \) does not provide a valid answer, then a singular solution does not exist.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

General Solution
In differential equations, a *general solution* is essentially a family of solutions that assumes various possible constant values. These solutions encompass all possible scenarios for a differential equation. When one looks at the problem, where the equation is given by \( p(xp - y + k) + a = 0 \), the idea is to find a relationship for \( y \) in terms of \( x \) that satisfies this equation for any constant \( c \). This constant \( c \) represents a whole set of possible values that the expression can take due to its arbitrary nature.

To solve for the general solution, we first substitute \( p = \frac{dy}{dx} \) into the equation. After simplification and assumption of \( y = cx + f(c) \), the derived general solution becomes \( y = cx + ck + \frac{a}{c} \). Here, \( c \) can be any real constant, thereby describing a multitude of solutions depending on the value of \( c \).

This solution allows for adaptability in specific cases and ensures coverage of every potential scenario the differential equation could represent.
  • The general solution represents a "generic" form adaptable to further specific initial conditions.
  • It describes a whole family of curves or potential solutions rather than a single pathway.
Singular Solution
A *singular solution* is a unique solution to a differential equation that cannot be derived from the general solution by choosing particular values of the constant \( c \). It represents an exceptional case that stands apart from the family of solutions given by the general solution. While many differential equations do not have a singular solution, it's important to check for their existence.

In the provided exercise, investigation for singular solutions requires differentiation of the original equation. By substituting \( p = 0 \) under special circumstances, we attempt to resolve if there exists a scenario where the equation has a distinct interpretation, separated from the general solution.

After performing these explorations, the check showed that if \( p = 0 \) does not provide a valid outcome, then a singular solution doesn’t emerge. Therefore, the absence of a singular solution in this instance means every case is adequately captured by variations of the general solution.
  • Singular solutions, if exist, offer unique solutions disconnected from the general family.
  • Existence checks are crucial as not every differential equation will present both types of solutions.
First Order Differential Equation
A *first order differential equation* involves derivatives of the first degree and no higher. In simpler terms, it includes derivatives like \( \frac{dy}{dx} \) but not \( \frac{d^2y}{dx^2} \) or higher. Understanding the order is essential because it dictates the types of methods one might employ to resolve the equation.

The equation given in the exercise, \( p(xp - y + k) + a = 0 \), is a clear indicator of a first order differential equation. Here, \( p = \frac{dy}{dx} \), making it first order since it deals directly with only the first derivative of \( y \) in respect to \( x \).

First order equations are often simpler to handle compared to higher-order ones and usually involve basic integration techniques or algebraic manipulations:
  • They require solving for a first derivative and integrating to find \( y \).
  • It's crucial to distinguish the order to apply correct solving strategies effectively.
Implicit Differentiation
*Implicit differentiation* is a technique used when functions are not expressed explicitly as \( y = f(x) \). Instead, they are intertwined with an equation where both \( x \) and \( y \) are mixed together, without an isolated function form. This technique is particularly valuable for handling more complex expressions.

In the problem instance, implicit differentiation plays a role when differentiating the original equation that mixes \( y \) and \( x \) through expressions involving their derivatives. This occurs during the exploration for singular solutions, where parts of the equation need to be differentiated with respect to \( x \). Implicit differentiation helps navigate these waters by respecting the relationship between \( y \) and \( x \) as dictated by the equation itself.
  • It allows handling of expressions where explicit expressions of functions are absent.
  • Vital for extracting derivative information in complex or non-standard forms.

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