Question

By the method of separation of variables solve \(y^{\prime}=(\cos x)(y-2)\) satisfying the initial condition \(y(0)=1\). Find the unique solution of the same equation satisfying \(y(0)=2 .\) Specify the interval on which the solution is defined in each case.

Short answer

Question: Find the unique solution of the differential equation \(y'(x)=(\cos x)(y-2)\) with initial conditions \(y(0)=1\) and \(y(0)=2\), and provide their defined intervals. Answer: The unique solution for the initial condition \(y(0)=1\) is \(y(x) = -e^{\sin x} + 2\), with an interval of \((-\infty, \infty)\). For the initial condition \(y(0)=2\), the unique solution is \(y(x) = 2\), with an interval of \((-\infty, \infty)\).

Step-by-step solution

Rewrite the differential equation using the method of separation of variables

First, we need to rewrite the given equation in the form \(y'(t) = f(t) g(y(t))\). We are given: \( y^{\prime}(x) = (\cos x)(y-2) \) We have \(f(x) = \cos x\) and \(g(y) = y-2\). With these functions, we can rewrite the given equation as: \( \frac{dy}{dx} = (\cos x)(y-2) \) Now we can separate the variables by dividing both sides by \((y-2)\) and multiplying both sides by \(dx\): \( \frac{dy}{y-2} = \cos x \ dx \)

Integrate both sides of the equation

Next, integrate both sides of the equation with respect to their respective variables: \( \int \frac{dy}{y-2} = \int \cos x \ dx \) For the left side, we have the integral of \(\frac{1}{y-2}\) with respect to \(y\). This integral is straightforward: \( \ln |y-2| = \int \cos x \ dx + C\) For the right side, we have the integral of \(\cos x\) with respect to \(x\). This integral is also straightforward: \( \ln |y-2| = \sin x + C \)

Solve for y(x)

Now, we will solve for y as a function of x by exponentiating both sides of the equation: \( y-2 = \pm e^{\sin x + C} \) \( y(x) = \pm e^{\sin x + C} + 2 \) This is the general solution of the differential equation.

Find the unique solution with the initial condition \(y(0)=1\)

Now we will use the initial condition \(y(0)=1\) to find the unique solution for this case: \( 1 = \pm e^{\sin 0 + C} + 2 \) Since \(\sin 0=0\) and \(e^0=1\), we have: \( 1 = \pm 1 + 2 \) This gives the unique solution: \( y(x) = -e^{\sin x} + 2 \) The interval for this solution is the entire real number line, \((-\infty, \infty)\).

Find the unique solution with the initial condition \(y(0)=2\)

Now we will use the initial condition \(y(0)=2\) to find the unique solution for this case: \( 2 = \pm e^{\sin 0 + C} + 2 \) This gives the unique solution: \( y(x) = 2 \) The interval for this solution is also the entire real number line, \((-\infty, \infty)\).

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are mathematical equations that involve functions of one independent variable and their derivatives. They are used to model a wide range of real-world phenomena, including physics, engineering, economics, and biology. An ODE typically looks like this: \( y'(x) = f(x, y) \), where \( y' \) denotes the derivative of \( y \) with respect to \( x \).
This simply means that the rate of change of \( y \) is defined by a function involving both \( x \) and \( y \) themselves.
One of the most important tasks in calculus is solving these ODEs, which provides insights into the behavior of the system being modeled. ODEs can be simple, involving a basic relationship between variables, or complex, requiring advanced mathematical techniques to find solutions.
In this exercise, the given equation is \( y^{\prime}(x) = (\cos x)(y-2) \). It builds upon a standard form where the derivative of \( y \) is described as a product of functions depending separately on \( x \) and \( y \).
The challenge is to find the function \( y(x) \) that satisfies this relationship.

Initial Value Problems

An Initial Value Problem (IVP) involves solving a differential equation along with a specified value at a particular point, often at the beginning of the domain, known as the initial condition. The purpose of an initial condition like \( y(0)=1 \) or \( y(0)=2 \) is to determine a unique solution from the general solution of an ODE.
Without an initial condition, a differential equation might have many possible solutions. The initial condition serves as a filter to find which one of these solutions applies to a specific situation.
In our exercise, we see two initial value problems. The first one with \( y(0)=1 \) helped find a unique function \( y(x) = -e^{\sin x} + 2 \), and an interval \((\-∞, ∞)\) where this solution is valid.
For the second IVP \( y(0)=2 \), the resulting solution is simply \( y(x) = 2 \), which is also valid over the same interval. Thus, initial conditions profoundly influence both the form of solutions and their applicability across intervals.

Integration Techniques

Integration is a critical step in solving differential equations, particularly when using the Separation of Variables method. The goal is to transform a differential equation into an integrable form with respect to each variable.
In our original problem, the separation of variables led to two integrals: one for \( y \) and another for \( x \). Specifically, we had:

• \( \int \frac{dy}{y-2} = \ln |y-2| \)
• \( \int \cos x \, dx = \sin x + C \)

The technique involves breaking down the differential into simpler parts that can be integrated using standard methods. These integrals yield expressions that, after solving for the dependent variable, provide a solution to the ODE.
Integration techniques such as substitution, integration by parts, or recognizing standard integral forms are essential tools. In the end, these integrations allowed us to solve \( \ln |y-2| = \sin x + C \) to find the general solution: \( y(x) = \pm e^{\sin x + C} + 2 \).
For each IVP case, integrating correctly and using the initial conditions are essential to finding the unique function that describes the situation accurately.