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Solving a Differential Equation In Exercises \(3-12\) , find the general solution of the differential equation. $$\frac{d y}{d x}=x+3$$

Short Answer

Expert verified
The general solution of the differential equation is \(y(x) = \frac{1}{2}x^2 + 3x + C\).

Step by step solution

01

Identification

Identify that the differential equation is separable which can be solved by direct integration. The equation here is \(\frac{d y}{d x}=x+3\) which clearly meets this criterion.
02

Integrating both sides

Perform the integration of both sides. We integrate the left side with respect to y and the right side with respect to x. This leads us to: \( \int dy = \int (x+3) dx\).
03

Compute the integrals

Since the integrals are straightforward, we have: \(y = \frac{1}{2}x^2 + 3x + C\) where \(C\) is the constant of integration.
04

Writing the general solution

The general solution of the differential equation is therefore \(y(x) = \frac{1}{2}x^2 + 3x + C\)

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Key Concepts

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

Separable Differential Equations
Understanding separable differential equations is foundational to grasping the nature of differential calculus. In essence, a separable differential equation is one that can be expressed in the form \(\frac{dy}{dx} = g(x)h(y)\), where the variables x and y can be separated on different sides of the equation. This allows us the luxury of treating 'dy' and 'dx' as individual entities, conveniently reorganizing the equation into \(h(y)dy = g(x)dx\).

The beauty of this method lies in its simplicity: by integrating both sides, we can find a general solution. The equation from our exercise, \(\frac{dy}{dx} = x + 3\), is an excellent example. Here, 'x' and 'y' are already separated, and we can integrate directly with respect to each variable.

In the context of your study, always first attempt to rearrange any given differential equation to see if it falls into this category of separable equations. This will often be your first step towards a solution.
Integration Methods

Techniques of Integration

Confronted with the task of integrating, you can pick from an arsenal of techniques suited for different functions. The aim is to find an antiderivative or integral of a given function. Among the common methods are:
  • Direct integration for elementary functions
  • Integration by parts
  • Substitution method
  • Partial fractions
  • Trigonometric substitution
Every method has its own specific use-case. For instance, in the provided exercise, we employed direct integration because the function was an elementary one, easily expressed as a sum of a polynomial and constant - the exact type of scenario where direct integration shines.

Understanding when and how to apply each method is paramount, as the choice of technique can dramatically simplify the problem. Looking at the integral of \(x+3\), a simple polynomial function, we can immediately integrate to \((x^2/2 + 3x + C)\), where \(C\) symbolizes the constant of integration. This is thanks to direct integration, underlining its value in solving differential equations.
General Solution of a Differential Equation
The general solution of a differential equation is essentially the 'complete' set of solutions that encompasses all possible particular solutions. It's a formula containing an arbitrary constant (or constants), which when given specific values, can describe all potential scenarios represented by the equation.

In the differential equation from our exercise, \(\frac{dy}{dx} = x + 3\), post-integration, we have \(y(x) = \frac{1}{2}x^2 + 3x + C\). Here, \(C\) is our constant of integration, symbolizing an infinite range of possible solutions that vary with this constant's value. As every different value of \(C\) gives us a new particular solution - a specific curve on the graph - the term 'general solution' encompasses all these curves.

It's essential to understand that obtaining the general solution is not the end. To fully resolve a problem, we often need additional information, such as initial conditions, to find a particular (unique) solution. This process, called 'solving an initial value problem', allows us to plug in initial conditions and solve for the constant, pinning down a single, definitive solution out of the infinite possibilities presented by the general solution.

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Most popular questions from this chapter

Population Growth When predicting population growth, demographers must consider birth and death rates as well as the net change caused by the difference between the rates of immigration and emigration. Let \(P\) be the population at time \(t\) and let \(N\) be the net increase per unit time resulting from the difference between immigration and emigration. So, the rate of growth of the population is given by $$\frac{d P}{d t}=k P+N$$ where \(N\) is constant. Solve this differential equation to find \(P\) as a function of time, when at time \(t=0\) the size of the population is \(P_{0} .\)

Population In Exercises \(51-54,\) the population (in millions) of a country in 2015 and the expected continuous annual rate of change \(k\) of the population are given. (Source: U.S. Census Bureau, International Data Base) (a) Find the exponential growth model \(P=C e^{k t}\) for the population by letting \(t=5\) correspond to \(2015 .\) (b) Use the model to predict the population of the country in \(2030 .\) (c) Discuss the relationship between the sign of \(k\) and the change in population for the country. \begin{array}{l}{\text { Country }}&{\text { 2015 Population }}&&{\text { \(k\) }} \\ {Paraguay}& {\text { 6.8}}&&{{0.012}}\end{array}

Describing Values Describe what the values of \(C\) and \(k\) represent in the exponential growth and decay model \(y=C e^{k t} .\)

Slope Field Describe the slope field for a logistic differential equation. Explain your reasoning.

In Exercises 47 and \(48,\) (a) use a graphing utility to graph the slope field for the differential equation, (b) find the particular solutions of the differential equation passing through the given points, and (c) use a graphing utility to graph the particular solutions on the slope field in part (a). \(\begin{array}{ll}{\text { Differential Equation }} & {\text { Points }} \\\ {\frac{d y}{d y}{d x}+4 x^{3} y=x^{3}} & \quad\left(0, \frac{7}{2}\right),\left(0,-\frac{1}{2}\right)\end{array}\)

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