Question

Solve the initial value problem $$ y^{\prime}=\left(2-e^{x}\right) /(3+2 y), \quad y(0)=0 $$ and determine where the solution attains its maximum value.

Short answer

Answer: The maximum value of the solution is attained at $x = \ln(2)$ and is equal to $y = \frac{4 - 3(\ln 2)}{2}$.

Step-by-step solution

Identify the Problem Type and Separate Variables

We are given the first-order, separable differential equation: $$ y^{\prime}=\frac{2-e^{x}}{3+2y} $$ Separate variables by taking the reciprocal on both sides and rewritting in terms of dy and dx: $$ \frac{dy}{dx} = \frac{3+2y}{2-e^x} \Longrightarrow (3+2y)dy = (2-e^x)dx $$

Integrate Both Sides

Integrate both sides of the equation with respect to their respective variables: $$ \int (3+2y)dy = \int (2-e^x)dx $$ Now, perform the integration: $$ \frac{1}{2}(3y + y^2) = 2x - e^x + C $$ Here, C is the constant of integration.

Apply the Initial Value

We are given that y(0) = 0. Plug this value into the equation to find the constant of integration: $$ \frac{1}{2}(3(0) + (0)^2) = 2(0) - e^0 + C \Longrightarrow 0 = 1 + C $$ So, the constant of integration is C = -1. Therefore, the specific solution is $$ \frac{1}{2}(3y + y^2) = 2x - e^x -1 $$

Determine the Maximum Value of the Solution

Taking the derivative of both sides, with respect to x: $$ \frac{1}{2}(3 + 2y)\frac{dy}{dx} = 2 - e^x $$ Now, we find the critical points by setting dy/dx = 0: $$ \frac{1}{2}(3 + 2y)(0) = 2 - e^x $$ Solve for x to get: $$ x = \ln 2 $$ Now, plug this x value into the specific solution we found in step 3: $$ \frac{1}{2}(3y + y^2) = 2(\ln 2) - e^{\ln 2} -1 $$ Solve for y to obtain the maximum value of the solution: $$ y = \frac{4 - 3(\ln 2)}{2} $$ So, the maximum value of the solution is attained at x = ln(2) and is equal to y = (4 - 3(ln 2))/2.

Key concepts

Separable Differential Equations

A differential equation is classified as 'separable' if the variables involved can be split onto either side of the equation, allowing each side to be functionally dependent on a single variable. In essence, this means that a separable differential equation can be written in the form \( \frac{dy}{dx} = g(x)h(y) \) where \( g(x) \) and \( h(y) \) are functions of \( x \) and \( y \) exclusively.
To solve such an equation, one would typically follow these steps: Firstly, rearrange the equation to express \( dy \) and \( dx \) in terms of \( g(x) \) and \( h(y) \). Secondly, integrate both sides independently. And finally, apply the initial conditions to find the particular solution. When working with the differential equation \( y'=(2-e^x)/(3+2y) \), it is manipulated into an appropriate form for integration, yielding \( (3+2y)dy = (2-e^x)dx \), demonstrating the variable separation clearly with \( y \) on one side and \( x \) on the other.

Integration Techniques

Solving a differential equation often requires integration to find the solution function. In this specific case, after separating the variables, we are left with two integrals: \( \int (3+2y)dy \) and \( \int (2-e^x)dx \). Integral calculus provides a variety of techniques to evaluate such expressions. Common methods include u-substitution, integration by parts, and recognizing standard integral forms.
For the first integral, the process is straightforward as the expression is already in a simple form; it involves the sum of a constant and a linear function of \( y \). For the second integral, we have an exponential function \( e^x \), which also has a well-known antiderivative. The respective antiderivatives yield the functions on both sides, and introduce an arbitrary constant \( C \) after integration which encapsulates all constants of integration. Subsequent simplification then provides a relation between \( x \) and \( y \).

Applying Initial Conditions

Initial value problems involve finding a specific solution to a differential equation that meets a given set of initial conditions. In the case at hand, we are given one such condition: \( y(0) = 0 \).
To apply this, substitute \( x = 0 \) and \( y = 0 \) into the general solution of the differential equation. This allows for the determination of the previously arbitrary constant \( C \), transforming the general solution into a particular one that satisfies the initial conditions. Once the constant is found, the solution to the initial value problem can be fully written out. By then taking the derivative of both sides and finding the value of \( x \) that causes the derivative to be zero, we can determine the maximum value of the solution, a common request in such problems. In this instance, after applying the initial conditions, we find the maximum value of the solution occurs at a specific point on the \( x \) axis, providing both the location and value of the solution's peak.