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Chapter 2: Problem 45

Use a technique of integration or a substitution to find an explicit solution of the given differential equation or initial value problem. $$\frac{d y}{d x}=\frac{1}{1+\sin x}$$

Short Answer

Expert verified
The explicit solution is: \[ y(x) = \tan(x/2) - \ln|1+\tan(x/2)| + C \]

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{1}{1+\sin x} \). This is a first-order ordinary differential equation where \( y \) is expressed in terms of \( x \). There is a separable form of the equation since \( y \) does not appear on the right side. This suggests that we can directly integrate with respect to \( x \).
02

Integrate Both Sides

We can integrate both sides of the equation with respect to \( x \). The left side becomes \( \int dy \), which integrates to \( y \), and on the right side, we have the integral \( \int \frac{1}{1+\sin x} \, dx \).
03

Use a Trigonometric Identity

To simplify the integral on the right, use the identity for \( 1 + \sin x = (\sin x/2 + \cos x/2)^2 \). This helps to substitute \( \sin x \rightarrow u \) for an easier integration. Alternatively, recognize that the integral can be simplified using a substitution trick or looking up a standard form.
04

Solve the Integral

The integral \( \int \frac{1}{1+\sin x} \, dx \) can be solved using the substitution \( t = \tan(x/2) \), where \( \sin x = \frac{2t}{1+t^2} \) and \( dx = \frac{2 \, dt}{1 + t^2} \). Through substitution and simplification, we find that: \[ \int \frac{1}{1+\sin x} \, dx = \int \frac{1+t^2}{(1+t)^2} \, dt \], and this simplifies to: \[ \int \left( 1 - \frac{1}{1+t} \right) \, dt \] \( \rightarrow t - \ln|1+t| + C \).
05

Back-substitute to Original Variable

Recall \( t = \tan(x/2) \), substitute back to get the explicit form in terms of \( x \): \[ y = \tan(x/2) - \ln|1+\tan(x/2)| + C \].
06

Write the Final Explicit Solution

Combining all terms in terms of \( x \), the explicit solution is: \[ y(x) = \tan(x/2) - \ln|1+\tan(x/2)| + C \], where \( C \) is the constant of integration. Ensure that all modifications have been reverted to express the solution in terms of \( x \).

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Key Concepts

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

Integration Techniques
Integration is a powerful technique used extensively in calculus, allowing us to find the function whose derivative is known. In this exercise, we applied integration to solve a first-order differential equation. There are various techniques for integration, each suited for different types of functions.

  • **Direct Integration**: Simple when the function is easily integrable with standard methods or known integration rules.
  • **Substitution Method**: Useful for complicated integrands, by changing variables to simplify the integral.
  • **Integration by Parts**: Utilized when the integrand is a product of functions.
In this specific problem, the function to integrate involved a trigonometric component, leading us to use substitution for simplicity.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) involve equations with functions and their derivatives concerning one independent variable, often denoted as 'x'. These equations help model real-world phenomena, such as motion or growth processes.

The exercise was focused on an ODE where the derivative of the function, \( \frac{dy}{dx} \), had to be integrated to find the explicit function \( y(x) \).

  • **Types of ODEs**: They can be 'first-order', 'higher-order', or 'linear' depending on the degree and form.
  • **Applications**: ODEs are used in physics, engineering, biology, and economics to represent systems and their dynamics.
In this problem, recognizing the order and type was crucial to selecting the right integration method.
First-Order Differential Equations
First-order differential equations involve derivatives of the first degree, that is, they involve the first derivative of the function (\( \frac{dy}{dx} \)). These are foundational in understanding how quantities change rate over time or space.

To solve these:
  • **Separate Variables**: If possible, express the equation so each side involves only one variable.
  • **Integrate Both Sides**: After separation, integrate both sides to find the solution.
  • **Apply Initial Conditions**: If given, these refine the solution to a particular path.
In our exercise, the separable nature allowed us to proceed by integrating each side with respect to \( x \). This leads to the explicit function \( y(x) \), completing the solution.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals that involve square roots or other expressions incorporating trigonometric functions. By substituting a trigonometric function for a variable, complex integrals can be transformed into simpler forms.

In our solution, the original integrand \( \frac{1}{1+\sin x} \) suggested this technique for simplification:
  • **Identify the Form**: Recognize if the integral fits a trigonometric identity or pattern.
  • **Use Substitution**: Substitute variables using identites like \( \tan(x/2) \) for \( x \), helping convert the integral.
  • **Back-substitute**: Replace substitutions with original variables to finalize the integration result.
This method allowed us to tackle the integration puzzle effectively, finally reducing it to a more straightforward algebraic expression.

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Most popular questions from this chapter

Explain why it is always possible to express any homogeneous differential equation \(M(x, y) d x+N(x, y) d y=0\) in the form \(\frac{d y}{d x}=F\left(\frac{y}{x}\right)\). You might start by proving that \(M(x, y)=x^{\alpha} M(1, y / x)\) and \(N(x, y)=x^{\alpha} N(1, y / x)\).

Find the general solution of the given differential equation. Give the largest interval \(I\) over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$\frac{d P}{d t}+2 t P=P+4 t-2$$

(a) Construct a linear first-order differential equation of the form \(x y^{\prime}+3 y=g(x)\) for which \(y=x^{3}+c / x^{3}\) is its general solution. Give an interval \(I\) of definition of this solution. (b) Give an initial condition \(y\left(x_{0}\right)=y_{0}\) for the DE found in part (a) so that the solution of the IVP is \(y=x^{3}-1 / x^{3}\). Repeat if the solution is \(y=x^{3}+2 / x^{3}\). Give an interval \(I\) of definition of each of these solutions. Graph the solution curves. Is there an initial-value problem whose solution is defined on \((-\infty, \infty) ?\) (c) Is each IVP found in part (b) unique? That is, can there be more than one IVP for which, say, \(y=x^{3}-1 / x^{3}, x\) in some interval \(I,\) is the solution?

Another population model is given by $$\frac{d P}{d t}=k P-h$$ where \(h\) and \(k\) are positive constants. For what initial values \(P(0)=P_{0}\) does this model predict that the population will go extinct?

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$y^{\prime}+4 x y=x^{3} e^{x^{2}}, \quad y(0)=-1$$
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