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Finding a General Solution Using Separation of Variables In Exercises \(1-14,\) find the general solution of the differential equation. $$ (2+x) y^{\prime}=3 y $$

Short Answer

Expert verified
The general solution to the differential equation is given by \(y = \sqrt{\frac{2}{3}(x^2 + 2x) - \frac{2}{3}C}\)

Step by step solution

01

Arrange the Given Equation

First arrange the equation to express it as \(y'(x) = G(x,y)\), we have: \[y'(x) = \frac{3y}{2+x}\]. We see that this function is separable meaning we can press variables on to one side each, let's do that.
02

Separate the Variables

We can express the above equation as \[(2+x) dx = 3y dy \] Now we integrate both sides of the equation: \[\int (2+x) dx = \int 3y dy \]
03

Integrate

When we integrate, we will get: \[x^2 + 2x = \frac{3}{2}y^2 + C\] Multiply both side of the above equation by \(\frac{2}{3}\), we get: \[\frac{2}{3}(x^2 + 2x) = y^2 + \frac{2}{3}C\]
04

Isolate y

Now isolate y to find the general solution: \[y = \sqrt{\frac{2}{3}(x^2 + 2x) - \frac{2}{3}C}\] Hence, the general solution is \(y = \sqrt{\frac{2}{3}(x^2 + 2x) - \frac{2}{3}C}\)

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Key Concepts

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

Differential Equations
A differential equation is a type of mathematical equation that involves an unknown function along with its derivatives. These equations play a critical role in various fields such as physics, engineering, and economics, as they describe various phenomena like motion, heat, or waves. In simple words, differential equations can be thought of as equations that describe change. For example, the differential equation in our exercise, \[ y' = \frac{3y}{2+x} \], describes how the function \( y \) changes with respect to \( x \).
  • "Differential" refers to the derivatives, which indicate rates of change.
  • "Equation" implies a mathematical expression that combines these derivatives.
Integration
Integration is the mathematical process of finding the integral of a function. It serves as the reverse operation of differentiation and is a vital tool in solving differential equations. When dealing with separable differential equations like the one we have, integration allows us to determine the function from its rate of change.
  • Integration helps in accumulating quantities, such as areas under curves.
  • It can also be used for determining antiderivatives in calculus.
In the given step-by-step solution, integration is utilized when both sides of the equation are separately integrated:\[ \int (2+x) \, dx = \int 3y \, dy \] This step is crucial in solving the equation and finding the unknown function \( y \).
General Solution
The general solution of a differential equation encompasses all possible solutions, including arbitrary constants that might arise during integration. In the exercise, the general solution is determined after integrating both sides of the equation and solving for \( y \).
The concept of a general solution is important because different initial conditions applied to the same differential equation might yield different specific solutions. Thus, having the general solution allows us to capture all such possibilities. The exercise concludes with the general solution:\[ y = \sqrt{\frac{2}{3}(x^2 + 2x) - \frac{2}{3}C} \] Here, \( C \) is the arbitrary constant, which will account for these variations.
Separable Equations
Separable equations are a special class of differential equations where the variables can be separated to opposite sides of the equation. This characteristic simplifies the process of finding a solution. Example of such an equation is:\[ y' = \frac{3y}{2+x} \]
  • We separate the variables by moving all terms involving \( y \) to one side and terms involving \( x \) to the other:
  • \( (2+x) \, dx = 3y \, dy \)
By simplifying and integrating each side, we can find the general solution of the equation. This method is powerful for solving first-order differential equations and is widely used in both academic and professional settings.

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Most popular questions from this chapter

Finding Orthogonal Trajectories In Exercises \(41-46,\) find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. $$ y^{2}=C x^{3} $$

Solving a Bernoulli Differential Equation In Exercises \(59-66,\) solve the Bernoulli differential equation. The Berchoulli equation is a well-known nonlinear equation of the form $$ y^{\prime}+P(x) y=Q(x) y^{n} $$ that can be reduced to a linear form by a substitution. The general solution of a Bernoulli equation is $$ y^{1-n} e^{\int(1-n) P(x) d x}=\int(1-n) Q(x) e^{f(1-n) P(x) d x} d x+C $$ $$ y^{\prime}+x y=x y^{-1} $$

Solving a Bernoulli Differential Equation In Exercises \(59-66,\) solve the Bernoulli differential equation. The Berchoulli equation is a well-known nonlinear equation of the form $$ y^{\prime}+P(x) y=Q(x) y^{n} $$ that can be reduced to a linear form by a substitution. The general solution of a Bernoulli equation is $$ y^{1-n} e^{\int(1-n) P(x) d x}=\int(1-n) Q(x) e^{f(1-n) P(x) d x} d x+C $$ $$ y^{\prime}+\left(\frac{1}{x}\right) y=x \sqrt{y} $$

Slope Field In Exercises 31 and \(32,\) sketch a few solutions of the differential equation on the slope field and then find the general solution analytically. To print an enlarged copy of the graph, go to MathGraphs.com. $$ \frac{d y}{d x}=x $$

lectric Circuits In Exercises 33 and \(34,\) use the differential equation for electric circuits given by \(L \frac{d I}{d t}+R I=E\) . In this equation, \(I\) is the current, \(R\) is the resistance, \(L\) is the inductance, and \(E\) is the electromotive force (voltage). Use the result of Exercise 33 to find the equation for the current when \(I(0)=0, E_{0}=120\) volts, \(R=600\) ohms, and \(L=4\) henrys. When does the current reach 90\(\%\) of its limiting value?

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