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Chapter 7: Problem 5

Solve the differential equation. $$( y + \sin y ) y ^ { \prime } = x + x ^ { 3 }$$

Short Answer

Expert verified
The solution is \( \frac{y^2}{2} - \cos y = \frac{x^2}{2} + \frac{x^4}{4} + C \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(( y + \sin y ) y' = x + x^3\). We notice it is a first-order differential equation.
02

Separate Variables

To separate the variables, rewrite the equation as \( (y + \sin y) dy = (x + x^3) dx \). This allows us to separate the equation with \( y \) terms on one side and \( x \) terms on the other.
03

Integrate Both Sides

Integrate both sides: \( \int (y + \sin y) dy = \int (x + x^3) dx \). The left side becomes \( \int y \, dy + \int \sin y \, dy = \frac{y^2}{2} - \cos y + C_1 \), and the right side becomes \( \frac{x^2}{2} + \frac{x^4}{4} + C_2 \).
04

Combine the Constants

Combine the constants from the integration, denoting the constant by \( C \): \( \frac{y^2}{2} - \cos y = \frac{x^2}{2} + \frac{x^4}{4} + C \).
05

Rearrange the Equation

Rearrange the equation to express the relationship between \( y \) and \( x \): \( \frac{y^2}{2} - \cos y - \frac{x^2}{2} - \frac{x^4}{4} = C \). This shows the integrated form of the solution.
06

Solve for Particular Solution (if initial conditions given)

If specific initial conditions are provided, substitute them into the equation \( \frac{y^2}{2} - \cos y = \frac{x^2}{2} + \frac{x^4}{4} + C \) to solve for the constant \( C \). If not, the general solution is sufficient.

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Key Concepts

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

First-Order Differential Equations
A first-order differential equation is characterized by involving the first derivative of the unknown function but not any higher derivatives. In our exercise, the differential equation \(( y + \sin y ) y' = x + x^3\) is a perfect example of this.
  • "First-order" means the highest derivative is the first derivative (\(y'\) or\(\frac{dy}{dx}\)).
  • They often express some kind of rate of change. In our case, the change of \(y\) with respect to \(x\).
  • Because they involve only one derivative, these equations are usually used in simpler physical situations where the rate of change is proportional to some driving force.
Recognizing the type of differential equation is important because it informs us what methods to use for solving. Here, since we have a first-order equation, we can use methods such as separation of variables, which we will discuss further.
Separation of Variables
Separation of variables is a powerful technique used to solve first-order differential equations like ours. The main idea is to rearrange the equation so that each variable and its differential is on a different side of the equation.
  • In our problem, we separate variables by rewriting the equation as \((y + \sin y) dy = (x + x^3) dx\).
  • This means we have all terms involving \(y\) on one side (left) and all terms involving \(x\) on the other side (right).
This separation allows us to perform integration on both sides easily, which is the next step in solving these types of equations. It's a little like unmixing two solutions, making it easier to see how changes in one variable affect changes in the other.
Integration Techniques
Once the variables are separated, we proceed by integrating both sides of the differential equation. Integration is essentially the reverse process of differentiation and it combines all small changes into a total amount.
  • For the equation \(\int (y + \sin y) dy = \int (x + x^3) dx\), we perform integration separately on each side.
  • The left integral \(\int y \, dy + \int \sin y \, dy\) results in \(\frac{y^2}{2} - \cos y\).
  • The right integral \(\int x \, dx + \int x^3 \, dx\) becomes \(\frac{x^2}{2} + \frac{x^4}{4}\).
After integrating, we get an equation involving an arbitrary constant \(C\) since indefinite integrals include a constant of integration. This constant accounts for any initial condition that might be given. If an initial condition is provided, it can be used at this stage to find a particular solution by determining the specific value of \(C\). In cases where no conditions are specified, the solution remains in its general form.

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Most popular questions from this chapter

\(29 - 31\) An integral equation is an equation that contains an unknown function \(y ( x )\) and an integral that involves \(y ( x )\) . Solve the given integral equation. [Hint: Use an initial condition obtained from the integral equation. \(]\) \(y ( x ) = 2 + \int _ { 2 } ^ { x } [ t - t y ( t ) ] d t\)

(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. \(x=3 t-5, \quad y=2 t+1\)

A food web Lynx eat snowshoe hares, and snowshoe hares eat woody plants, such as willows. Suppose that, in the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially. In the absence of lynx and willow, the hare population will decay exponentially. If \(L(t), H(t),\) and \(W(t)\) represent the populations of these three species at time \(t\) write a system of differential equations as a model for their dynamics. If the constants in your equation are all positive, explain why you have used plus or minus signs.

Suppose that the population dynamics of a species obeys a modified version of the logistic differential equation having the following form: \(\frac{d N}{d t}=r\left(1-\frac{N}{K}\right)^{2} N\) where \(r \neq 0\) and \(K>0\) (a) Show that \(\hat{N}=0\) and \(\hat{N}=K\) are equilibria. (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (c) Apply the local stability criterion to the equilibrium \(\hat{N}=K .\) What do you think your answer means about the stability of this equilibrium? (Note: This is an example in which the local stability criterion is inconclusive.) (d) Construct two phase plots, one for the case where \(r>0\) and the other for \(r<0,\) and determine the stability of \(\hat{N}=K\) in each case. Does the answer match your reasoning in part \((c) ?\)

Show that \(y=\frac{2}{3} e^{x}+e^{-2 x}\) is a solution of the differential equation \(y^{\prime}+2 y=2 e^{x} .\) Is this differential equation pure-time, autonomous, or nonautonomous?
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