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Find the general solution of each homogeneous equation. $$ (x+y) d x+(y-x) d y=0 $$

Short Answer

Expert verified
Use substitution and integration; the general solution relates \( x(e^{\frac{y^2}{2x^2}}) \) to \( C(1 + \frac{y}{x}) \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation \[(x+y) \, dx + (y-x) \, dy = 0\]is a first-order homogeneous differential equation.
02

Check Homogeneity

To check homogeneity, we substitute \( x = kx \) and \( y = ky \) into the equation. The equation forms \[(kx + ky) \, dx + (ky - kx) \, dy = 0,\]which simplifies to \[k(x + y) \, dx + k(y - x) \, dy = 0.\]Both terms are of degree one, confirming that the equation is homogeneous.
03

Substitution and Simplification

Use the substitution \( v = \frac{y}{x} \), which gives \( y = vx \) and \( dy = v \, dx + x \, dv \). Substitute into the equation:\[(x + vx) \, dx + (vx - x) (v \, dx + x \, dv) = 0.\]This simplifies to:\[[1 + v] \, dx + [v^2 - 1] \, x \, dv = 0.\]
04

Separate Variables

Rearrange and separate variables:\[\frac{dx}{x} = -\frac{v^2 - 1}{1 + v} \, dv.\]
05

Integrate Both Sides

Integrate both sides:\[\int \frac{1}{x} \, dx = \int -\frac{v^2 - 1}{1 + v} \, dv.\]This results in:\[\ln|x| = -\int(v - \frac{1}{1+v}) \, dv.\]Integrate the right-hand side:\[-\int v \, dv + \int \frac{1}{1+v} \, dv = -\frac{v^2}{2} + \ln|1+v| + C.\]
06

Write the General Solution

Combine results to get:\[\ln|x| = -\frac{v^2}{2} + \ln|1+v| + C.\]Convert back to original variables using \( v = \frac{y}{x} \):\[|x| = \frac{|1 + \frac{y}{x}|}{e^{\frac{v^2}{2}}}\]This forms the basis of the general solution:\[x \, e^{\frac{y^2}{2x^2}} = C(1 + \frac{y}{x})\] or equivalent forms.

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Key Concepts

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

First-order Differential Equation
A first-order differential equation involves derivatives of the first degree, meaning it includes terms such as \( dx \) and \( dy \) but not higher-order derivatives like \( d^2x \) or \( d^2y \). It is pivotal because it simplifies our approach to solving problems involving rate and change.
First-order differential equations usually have the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \), where \( M \) and \( N \) are continuous functions of \( x \) and \( y \).
In the given exercise, you encounter a first-order differential equation, indicated by the presence of \( dx \) and \( dy \) without derivatives of higher order. This type of equation forms the basis for analyzing many natural phenomena, including motion and growth.
Variable Substitution
Variable substitution is a powerful method used to simplify differential equations by introducing a new variable. In the context of homogeneous equations, we often use the substitution \( v = \frac{y}{x} \).
This substitution translates the dependent and independent variables \( x \) and \( y \) into a single variable \( v \), turning a two-variable problem into a simpler form. Let's imagine you're standing on the shoulders of this idea, allowing the differential equation to be easier to crack.
  • With substitution, \( y = vx \).
  • Also, differentiate to find \( dy = v \, dx + x \, dv \).
Using these relations transforms the original equation into one purely involving \( v \) and \( x \), thereby smoothing the path toward integration.
Separation of Variables
Separation of variables is a technique for solving differential equations where variables are rearranged to facilitate integration. This approach is particularly useful when the equation can be expressed so that each side contains only one of the original variables.
After applying the substitution in our example, the equation is reduced into a manageable form, allowing us to separate \( dx \) and \( dv \). Thus, transforming:
  • \(\frac{dx}{x} = -\frac{v^2 - 1}{1 + v} \, dv \)
This separation aligns each side with a single variable, leading toward integration. It lays the groundwork for elucidating the intricate behaviors entailed by the differential equation.
Integration of Differential Equations
Integration is the process of solving a differential equation by finding the antiderivatives. After separation of variables, this process helps us unite the components of the differential equation.
In the exercise, we integrate both sides after separation:
  • \(\int \frac{1}{x} \, dx = \int -\frac{v^2 - 1}{1 + v} \, dv \)
This yields the solution in terms of natural logarithms and algebraic expressions, reflecting the accumulated effect over continuous movement.
Integration culminates in expressing the original problem within a comprehensible framework, revealing the general solution: \[ x \, e^{\frac{y^2}{2x^2}} = C(1 + \frac{y}{x}) \] This general solution encapsulates the essence of the differential equation, highlighting the beauty and application of calculus.

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Most popular questions from this chapter

Sensitivity to initial conditions is well illustrated by a little target practice with your numerical solver. In Exercises \(1-12\), you are given a differential equation \(x^{\prime}=f(t, x)\) and a "target." In cach case, enter the equation into your numerical solver, and then experiment with initial conditions at the given value of \(t_{0}\) until the solution of \(x^{\prime}=f(t, x)\), with \(x\left(t_{0}, x_{0}\right)\), "hits" the given target. We will use the simple linear equation, \(x^{\prime}=x-t\). The initial conditions are at \(t_{0}=0\). The target is $$ (6,0) $$

An autonomous differential equation is given in the form \(y^{\prime}=f(y)\). Perform each of the following tasks without the aid of technology. (i) Sketch a graph of \(f(y)\). (ii) Use the graph of \(f\) to develop a phase line for the autonomous equation. Classify each equilibrium point as either unstable or asymptotically stable. (iii) Sketch the equilibrium solutions in the \(t y\)-plane. These equilibrium solutions divide the ty-plane into regions. Sketch at least one solution trajectory in each of these regions. $$ y^{\prime}=(y+1)\left(y^{2}-9\right) $$

An electric circuit, consisting of a capacitor, resistor, and an electromotive force can be modeled by the differential equation $$ R \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ where \(R\) and \(C\) are constants (resistance and capacitance) and \(q=q(t)\) is the amount of charge on the capacitor at time \(t\). For simplicity in the following analysis, let \(R=C=1\), forming the differential equation \(d q / d t+q=E(t)\). In Exercises 17-20, an electromotive force is given in piecewise form, a favorite among engineers. Assume that the initial charge on the capacitor is zero \([q(0)=0]\). (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval \([0,4]\). (ii) Find an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period. E(t)= \begin{cases}5, & \text { if } 0

Use the variation of parameters technique to find the general solution of the given differential equation. Then find the particular solution satisfying the given initial condition. $$ y^{\prime}-3 y=4, \quad y(0)=2 $$

An autonomous differential equation is given in the form \(y^{\prime}=f(y)\). Perform each of the following tasks without the aid of technology. (i) Sketch a graph of \(f(y)\). (ii) Use the graph of \(f\) to develop a phase line for the autonomous equation. Classify each equilibrium point as either unstable or asymptotically stable. (iii) Sketch the equilibrium solutions in the \(t y\)-plane. These equilibrium solutions divide the ty-plane into regions. Sketch at least one solution trajectory in each of these regions. $$ y^{\prime}=(y+1)(y-4) $$

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