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Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(x)=y \cos x, y(0)=3$$

Short Answer

Expert verified
Question: Determine the particular solution of the initial value problem \(y^{\prime}(x) = y \cos x\) with the initial condition \(y(0)=3\). Answer: The particular solution for the given initial value problem is \(y(x) = 3 e^{\sin x}\).

Step by step solution

01

Determine if the equation is separable

A differential equation is separable if it can be written in the form: \(\frac{dy}{dx} = f(x)g(y)\). If the given equation \(y^{\prime}(x) = y \cos x\) is separable, then we should be able to rewrite it in the required form. Let \(f(x) = \cos x\) and \(g(y) = y\). Therefore, the given equation is indeed separable.
02

Rewrite the equation in terms of separate variables

To rewrite the equation in terms of separate variables, we need to divide both sides by \(dy\) and \(g(y)\) such that: $$\frac{dy}{y} = f(x) dx$$$$\frac{dy}{y} = \cos x dx$$
03

Integrate both sides

Now, we need to integrate both sides of the equation with respect to their respective variables: $$\int \frac{dy}{y} = \int \cos x dx$$$$\ln |y| = \sin x + C_{1}$$Where \(C_{1}\) is the integration constant.
04

Solve for y(x)

To determine y(x), take the exponent of both sides of the equation: $$y(x) = e^{\sin x + C_{1}}$$$$y(x) = e^{\sin x} e^{C_{1}}$$We can denote \(e^{C_{1}}\) as another constant \(C\), so the equation will look like:$$y(x) = C e^{\sin x}$$
05

Use the initial value \(y(0)=3\) to find the constant C

Now that we have a general solution for y(x), we can apply the initial value \(y(0)=3\) to find the constant C.$$3 = C e^{\sin (0)}$$$$3 = C e^{0}$$$$3 = C$$Thus, the particular solution of the initial value problem is:$$y(x) = 3 e^{\sin x}$$

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Key Concepts

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

Separable Differential Equation
A differential equation is termed 'separable' when it can be expressed as the product of functions, each depending solely on one variable—either independent or dependent. For example, the equation \(y'(x) = y \cos x\) is separable because it can be written in the form \(\frac{dy}{dx} = f(x)g(y)\), where \(f(x) = \cos x\) and \(g(y) = y\).

To solve a separable equation, you rearrange it so that all terms involving one variable, such as \(y\), appear on one side of the equation and all terms involving another variable, like \(x\), appear on the other. This technique ensures each side only contains one variable, allowing us to integrate both sides separately.
  • Identify functions \(f(x)\) and \(g(y)\).
  • Rewrite differential as \(\frac{1}{g(y)} dy = f(x) dx\).
  • Perform integration on both sides to progress towards a solution.
Recognizing and manipulating separable equations is fundamental in solving many initial value problems. It allows the equation to be split and solved step-by-step using straightforward integration.
Integration
Integration is a powerful mathematical process used to calculate accumulated quantities, like areas under curves or the antiderivatives of functions. In the context of differential equations, integration helps us find functions whose derivative matches the given differential equation.

When we separate the variables in a differential equation, each side of the equation is set up to be integrated individually. For example, in the problem \(\frac{dy}{y} = \cos x dx\), we need to integrate each side:
  • Integrate \(\int \frac{dy}{y}\) to find \(\ln|y|\).
  • Integrate \(\int \cos x dx\) to find \(\sin x\).
After integrating, you will often introduce a constant of integration to account for any constant that might have disappeared during differentiation. This process turns the isolated derivatives into actual functions, one step closer to finding the particular solution.
Particular Solution
A particular solution of a differential equation is a specific solution derived from the general solution by applying given conditions or constraints, often initial values. In this context, the initial value provided helps pinpoint one unique solution out of the infinite possibilities.

After finding the general solution, such as \(y(x) = Ce^{\sin x}\), you can use the initial value \(y(0) = 3\) to compute \(C\).
  • Substitute \(x = 0\) and \(y = 3\) into the equation.
  • Solve for \(C\), resulting in \(C = 3\).
This procedure narrows down the solution to one that satisfies the initial condition, yielding \(y(x) = 3e^{\sin x}\) as the particular solution. This approach ensures the solution not only solves the differential equation but also aligns with specified initial conditions.
Constant of Integration
The constant of integration is an essential concept in calculus representing any constant added to the antiderivative during integration. It emerges because differentiation obliterates constants, meaning when we work backward through integration, we must account for any original constant that might have existed.

In our problem, once each side of the separated differential equation was integrated, the results included \( \ln |y| = \sin x + C_1 \). Here, \(C_1\) is an arbitrary constant introduced during integration. This constant signifies the family of curves that satisfy our differential equation.
  • Constant arises in indefinite integration.
  • Represents an infinite number of potential solutions.
When dealing with initial value problems, exact values are often given, enabling us to address this "degree of freedom" and find a unique solution. The constant's value is determined by applying initial or boundary conditions as shown by setting \(y(0) = 3\) to solve for \(C\), which led to the particular solution.

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Most popular questions from this chapter

The amount of drug in the blood of a patient (in milligrams) administered via an intravenous line is governed by the initial value problem \(y^{\prime}(t)=-0.02 y+3\) \(y(0)=0,\) where \(t\) is measured in hours. a. Find and graph the solution of the initial value problem. b. What is the steady-state level of the drug? c. When does the drug level reach \(90 \%\) of the steady-state value?

What is a carrying capacity? Mathematically, how does it appear on the graph of a population function?

Consider the differential equation \(y^{\prime \prime}(t)-k^{2} y(t)=0,\) where \(k>0\) is a real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} e^{t}+C_{2} e^{-t} .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} e^{2 t}+C_{2} e^{-2 t}\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture. d. For a positive real number \(k\), verify that the general solution of the equation may also be expressed in the form \(y(t)=C_{1} \cosh k t+C_{2} \sinh k t,\) where cosh and sinh are the hyperbolic cosine and hyperbolic sine, respectively (Section \(7.3)\)

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\) b. Graph the solution with \(b=0.1 \mathrm{s}^{-1}\) c. Using the graph in part (b), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\)

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