Question

\(\left(2 e^{y}-x\right) y^{\prime}=1, y(0)=0,[-5,5]\). Hint: dsolve \(\left({ }^{\prime}(2 * \exp (\mathrm{y})-\mathrm{x}) * \mathrm{Dy}=1^{\prime}, \mathrm{y}^{\prime}(0)=0^{\prime}, \mathrm{x}^{\prime}\right)\).

Short answer

General solution: \(2e^y = x + 2\).

Step-by-step solution

Identify the Differential Equation

The given differential equation is \((2e^y - x)y' = 1\) with the initial condition \(y(0) = 0\). This is a first-order differential equation.

Solve for the Differential Equation

We start by rewriting the equation in a separable form. First, express \(y'\) as \( dy/dx \):\[ (2e^y - x) \frac{dy}{dx} = 1 \]. Rearrange to separate variables:\[ (2e^y - x) dy = dx \].

Integrate Both Sides

Integrate both sides with respect to their respective variables:\[ \int (2e^y) dy - \int x dy = \int dx \]. This yields:\[ 2e^y - x = x + C \] (where \(C\) is the integration constant).

Apply Initial Condition

Use the initial condition \(y(0) = 0\) to find \(C\):\(2e^0 - 0 = 0 + C \).We get \(2 = C\).

Formulate the General Solution

Now substitute \(C = 2\) back into the integrated equation:\[ 2e^y - x = x + 2 \]. This is our general solution.

Key concepts

Separable Equations

Separable equations are a specific type of differential equations that can be rewritten so each variable appears on a separate side of the equation. This simplification allows us to integrate each side independently. In this exercise, the differential equation is given by

• \((2e^y - x) y' = 1\)

To separate the variables, we rearrange it into:

• \((2e^y - x) dy = dx\)

Now, each side of the equation involves only one of the variables. The left side has all the terms involving \(y\), and the right side involves only \(x\).
This separation is crucial as it allows us to tackle each part with integration, easing the solving process.
This approach is particularly useful when working with first-order differential equations. It simplifies complex equations, making them easier to handle step by step.

Initial Value Problems

Initial value problems involve differential equations with an additional condition, known as the initial condition. These conditions specify the value of the unknown function at a particular point, helping to find a specific solution. In our case:

• The initial condition is \(y(0) = 0\).

Once we have integrated both sides, the integration constant \(C\) arises.
We use the initial condition to determine \(C\).For example, substituting the initial value:

• \(2e^0 - 0 = 0 + C\)
• \(2 = C\)

This step is vital because it adjusts the general solution of the differential equation to reflect the specific scenario described by the initial condition. Including the initial condition ensures that the solution is not just general, but uniquely meets the problem's requirements.

Integration Techniques

Integration is an essential skill in solving separable differential equations. Once the equation is separated, each side is integrated concerning its variable. In this exercise, we have:

• \(\int (2e^y) \, dy - \int x \, dy = \int dx\)

Using basic integration rules, we integrate:

• The left side yields \(2e^y\).
• The right side yields \(x + C\), where \(C\) is the constant of integration.

Integration not only helps find the antiderivatives but also plays a pivotal role when combined with initial conditions. This combination is what transforms a general solution into a specific one.
Mastering integration techniques allows you to navigate through different functions and solve various differential equations efficiently.