91Ó°ÊÓ

Separable differential equations are a special type where you can rewrite the equation so that each variable appears on a different side of the equation. This kind of equation typically looks like this: \( \frac{d y}{d x} = g(y) \cdot h(x) \). The goal is to rearrange it in a way that separates the variables, often looking like \( g(y) \cdot d y = h(x) \cdot d x \). The beauty of separable equations lies in their simplicity — when each side of the equation depends only on one variable, it becomes easier to solve by integrating.

• Identify the equation as separable: If you can rewrite it into a product of functions of \(x\) and \(y\).
• Separate the variables: Arrange the equation to isolate each variable and its differential on opposite sides.
• Integrate: Once separated, each side can be integrated independently with respect to its own variable.

In our problem, we started with \( \frac{d y}{d x} = e^{y} \), which was separable into \( \frac{d y}{e^{y}} = d x \). This step is crucial as it simplifies how we handle the equation.

Integration is at the core of solving differential equations. Once a separable equation is rearranged, you need to integrate both sides. This process transforms the problem from one of differentiation into one of finding antiderivatives.For the equation \( \frac{d y}{e^{y}} = d x \):

• Integrate both sides: This breaks down to solving two integrals: \( \int \frac{d y}{e^{y}} = -e^{-y} + C \) and \( \int d x = x + C' \).
• Combine the constants: Usually, you combine constants of integration from both sides into a single constant \( C \).

After integration, the equation becomes \( -e^{-y} = x + C \). This form is key because it represents the general solution of the differential equation, expressed implicitly in terms of \(x\) and \(y\). Solving for \(y\), we get \( y = -\ln(-x - C) \), showing the function form.

Initial conditions are essential for determining the particular solution of a differential equation. Without them, our solution would remain a family of curves defined by arbitrary constants.In this exercise, we were given the initial condition \( y(0) = 1 \). This condition means that when \( x = 0\), \( y \) must be equal to 1, providing a point that the solution curve must pass through.

• Substitute initial conditions: Substitute \( x = 0\) and \( y = 1 \) into the equation obtained after integration.
• Solve for the constant: This process yielded \( 1 = -\ln(-0 - C) \), which simplifies to \( e^{-1} = -C \). Thus, \( C = -e^{-1} \).

Finally, incorporating \( C \) back into the equation gives us the specific solution: \( y = -\ln(-x + e^{-1}) \). The initial conditions help pin down the exact curve from the infinite possibilities, ensuring the solution meets exact criteria.