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Chapter 9: Problem 18

Solve the initial-value problem. $$\cos y d x+\left(1+e^{-x}\right) \sin y d y=0, \quad y(0)=\pi / 4.$$

Short Answer

Expert verified
The given initial-value problem is: \[\cos y \, dx + (1 + e^{-x})\sin y \, dy = 0, \quad y(0)=\pi / 4.\] It was attempted to convert it into a standard first-order linear ordinary differential equation (ODE) and solve using an integrating factor. However, the integral for the integrating factor cannot be calculated directly using elementary functions, as it depends on both \(x\) and \(y\). Therefore, the problem cannot be solved using this approach.

Step by step solution

01

Identify the type of differential equation

The given differential equation is: \[\cos y \, dx + (1 + e^{-x})\sin y \, dy = 0\] The above equation is in the form of an exact differential equation. It can be written as: \[M(x,y) \, dx + N(x, y) \, dy = 0\] where: \(M(x,y)=\cos y\) \(N(x,y)=(1+e^{-x})\sin y\)
02

Check if the differential equation is exact

We will verify if the given equation is exact by comparing the partial derivatives of \(M\) and \(N\) with respect to the other variable. In other words, we need to check if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). The partial derivative of \(M\) with respect to \(y\): \[\frac{\partial M}{\partial y} = -\sin y\] The partial derivative of \(N\) with respect to \(x\): \[\frac{\partial N}{\partial x} = (-e^{-x})(\sin y) = - e^{-x} \sin y\] As the partial derivatives are not equal, we conclude that the given equation is not exact. Therefore, we cannot directly apply the method for exact differential equations.
03

Convert the differential equation into a standard form

As we cannot apply the method for solving exact differential equations, we will try to convert the given equation into a standard form. Dividing the equation by \(\cos y \, dy\), we get: \[\frac{dx}{dy}+P(x)\frac{dy}{dx} Q(y) = 0\] where: \(P(x) = \frac{1+e^{-x}}{\cos y}\) \(Q(y) = \sin y\) We can rewrite the equation as a first-order linear ordinary differential equation (ODE): \[\frac{dx}{dy}+P(x)Q(y)=0\]
04

Solve the first-order linear ODE

Now we have the ODE in the form \(\frac{dx}{dy} + P(x)Q(y) = 0\). We can use an integrating factor, \(\mu\), to solve this ODE. The integrating factor is defined as \(\mu(x) = e^{\int P(x) \, dx}\). In our case, \(P(x) = \frac{1+e^{-x}}{\cos y}\), so the integrating factor \(\mu(x)\) is given by: \[\mu(x) = e^{\int \frac{1+e^{-x}}{\cos y} \, dx}\] This integral cannot be calculated directly, as it depends on both \(x\) and \(y\). As a result, the given initial-value problem cannot be solved using elementary functions.

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Key Concepts

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

Exact Differential Equations
Exact differential equations are a special kind of differential equations. They have a specific form. It is \[ M(x, y) \, dx + N(x, y) \, dy = 0 \]where functions \( M(x, y) \) and \( N(x, y) \) are related through partial derivatives. The key condition for exactness is that the partial derivative of \( M \) with respect to \( y \) equals the partial derivative of \( N \) with respect to \( x \).
This means:
  • \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)
If this condition holds, the equation can be solved directly by integrating \( M \) with respect to \( x \) and \( N \) with respect to \( y \). However, sometimes the equation isn't exact initially. In such cases, we need to explore other methods to solve it.
First-order Linear ODE
First-order linear ordinary differential equations (ODEs) involve first derivatives and are of the form:\[ \frac{dy}{dx} + P(x)y = Q(x) \]These equations have a linear structure regarding the dependent variable and its derivative.
They are called "linear" because neither the function nor its derivative are raised to any power other than the first.To solve a first-order linear ODE, we apply a process that may include:
  • Rearranging terms to match the standard form
  • Using specific techniques like integrating factors
  • Transforming the equation if necessary
If the equation involves variables like \( y \) or both \( x \) and \( y \) in complex ways, finding solutions might require other approaches like numerical or approximate solutions.
Integrating Factor
The integrating factor is a powerful tool to solve first-order linear ODEs. It allows us to simplify differential equations to make them more solvable. Typically, the integrating factor is a function, denoted by \( \mu \), given by:\[ \mu(x) = e^{\int P(x) \, dx} \]
Applying the integrating factor involves:
  • Multiplying the entire equation by \( \mu(x) \)
  • Resulting in a form where the left side becomes the derivative of a product
  • Simplifying it, and then integrating both sides
However, integrating factors can sometimes be hard to find, especially when the function involves non-linearities with solutions that are not expressible through basic elementary functions.
Non-solvable by Elementary Functions
In mathematics, many differential equations can be solved in terms of elementary functions. These include functions like polynomials, exponentials, trigonometric, and logarithmic functions.
"Non-solvable by elementary functions" refers to cases where solutions cannot be expressed using such functions alone. This often happens when:
  • Complex relationships between variables exist
  • Integrals lead to functions outside the scope of basic elements
  • No straightforward or algebraic method works to express the solution
In such instances, solutions might involve special functions or require numerical methods to understand the behavior of the solution comprehensively.

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Most popular questions from this chapter

Find the general solution. $$2 y^{\prime \prime}+3 y^{\prime}=0$$

Find the integral curves. If the curves are the graphs of functions \(y=f(x)\). determine all the functions that satisfy the equation. $$y^{\prime}=3 x^{2}\left(1+y^{2}\right)$$.

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Determine whether the functions satisfy the deferential equation. $$\begin{array}{c} y^{\prime \prime}-2 y^{\prime}-3 y=7 e^{3 x}: \quad y_{1}(x)=e^{-x}+2 e^{3 x} \\\ y_{2}(x)=\frac{7}{3} x e^{3} \end{array}$$
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