91Ó°ÊÓ

An initial value problem is a type of differential equation accompanied by a specified value, known as the initial condition. This condition enables us to find a definite solution to the differential equation.
In the given exercise, the differential equation is \( \frac{d y}{d x} = x e^{y} \), and the initial condition provided is \( y(0) = 0 \).
The initial condition plays a critical role as it specifies the value of the function at a particular point, helping us to find the specific solution in contrast to general solutions that include arbitrary constants.

• Initial conditions often take the form \( y(x_0) = y_0 \), with \(x_0\) being the initial value of \(x\).
• The solutions to these problems give us unique functions that fulfill both the equation and the initial condition.

Without applying the initial value, we would only have a family of solutions represented by a constant \(C\). The initial condition helps calculate \(C\), thus determining the particular solution.

Separation of variables is a powerful technique for solving differential equations, especially when they can be rearranged to separate the variables. This technique is used to transform a complex differential equation into simpler integrals that are easier to solve.
In the exercise, we start with the equation \( \frac{d y}{d x} = x e^y \) and aim to rearrange it. By dividing both sides by \( e^y \) and multiplying by \( dx \), we obtain: \( e^{-y} dy = x dx \).

• The left side involves only \( y \) and its derivative \( dy \), and the right side involves only \( x \) and its derivative \( dx \).
• This separation allows us to tackle each side independently through integration.

The goal is to convert the original equation into two simpler integrals, making the problem easier to handle. Once the variables are separated, it sets the stage for integration, which will allow us to find the function \( y \).

Integration is a fundamental process in calculus used to find functions when the derivative is given. After separating the variables in a differential equation, each side is integrated with respect to its variable.
For our problem, the differential equation \( e^{-y} dy = x dx \) is split into two integrals: \( \int e^{-y} dy \) and \( \int x dx \).

• The solution of \( \int e^{-y} dy \) is \(-e^{-y}\), which arises from recognizing it as an exponential function.
• The solution of \( \int x dx \) is \( \frac{x^2}{2} + C \).

Here, \(C\) represents the constant of integration, and it's necessary because the indefinite integral yields a family of functions.
After integrating, we use the initial condition to solve for \(C\), ensuring that we find a particular solution that satisfies the given initial value.

A particular solution of a differential equation is the specific solution that satisfies the initial condition or conditions of the problem. This is in contrast to the general solution, which includes a constant of integration.
For our exercise, we started with the equation \(-e^{-y} = \frac{x^2}{2} + C\), after integration. Our goal was to find \( C \) using the initial condition \( y(0) = 0 \). Substituting \( x = 0 \) and \( y = 0 \) into the equation, we determined that \( C = -1 \).

• Substituting the value of \( C \) into the equation gives us \(-e^{-y} = \frac{x^2}{2} - 1 \).
• This leads to solving for \( y \), which ultimately renders the solution \( y = -\ln\left(1 - \frac{x^2}{2}\right) \).

The particular solution stands out because it's tailored to match the initial condition provided.
This solution not only satisfies the differential equation but agrees with \( y(0) = 0 \), demonstrating its accuracy as a particular solution.