Question

Find the solution to the initial-value problem.\(y^{\prime}=e^{y-x}, y(0)=0\)

Short answer

The solution to the initial-value problem is \( y = x \).

Step-by-step solution

Setup the Differential Equation

The given differential equation is \( y' = e^{y-x} \). We need to solve this with the initial condition \( y(0) = 0 \).

Separate Variables

First, rewrite the equation in a form that allows you to separate variables: \( y' = \frac{dy}{dx} = e^{y-x} \). This can be written as \( \frac{dy}{e^y} = e^{-x} dx \).

Integrate Both Sides

Integrate both sides to find \( y \). The left side integral becomes \( \int e^{-y} \, dy \), which evaluates to \( -e^{-y} \). The right side integral becomes \( \int e^{-x} \, dx \), which evaluates to \( -e^{-x} \). Thus, we have:\[-e^{-y} = -e^{-x} + C\],where \( C \) is the constant of integration.

Apply Initial Condition

Apply the initial condition \( y(0) = 0 \). Substituting into the integrated equation:\[-e^{0} = -e^{0} + C\] or \(-1 = -1 + C\),which simplifies to \( C = 0 \).

Solve for \( y \)

Using the result from the previous step, our equation simplifies to:\(-e^{-y} = -e^{-x}\). Solving for \( y \), we get:\[e^{-y} = e^{-x}\]\[-y = -x\]\( y = x \). Therefore, the solution is \( y = x \).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation problem that includes an initial condition.
This means the solution to the differential equation must not only satisfy the equation itself but also match a specific value at a particular point.
In this context, we are dealing with the initial value problem of the differential equation:

• Given differential equation: \( y' = e^{y-x} \)
• Initial condition: \( y(0) = 0 \)

The initial condition \( y(0) = 0 \) provides a specific point on the curve of the solution, essentially anchoring it. By applying this condition after integrating the separated variables, we determine the particular solution that satisfies both the differential equation and the initial condition.
This approach helps transform the general solution, which contains a constant of integration, into a specific function satisfying the initial parameters.

Separation of Variables

Separation of variables is a popular method used to solve ordinary differential equations, particularly when the equation can be algebraically manipulated to segregate the dependent and independent variables.
It relies on expressing the equation in a form where all terms involving the dependent variable \( y \) can be moved to one side, and all terms involving the independent variable \( x \) to the other side, making it possible to integrate both sides separately. In our example, starting with:

• \( y' = e^{y-x} \)

We rewrite this to segregate \( y \) and \( x \):

• \( \frac{dy}{e^y} = e^{-x} \, dx \)

This process effectively 'separates' the variables. From here, you can independently integrate each side with respect to its respective variable. This step is fundamental because it allows for the use of simple integration techniques to find solutions.

Integration

Integration is a mathematical operation that, loosely speaking, is the reverse of differentiation.
In solving differential equations, integration allows us to retrieve a function from its derivative.Once we have successfully separated the variables, the next step involves integrating both sides.
For the separated equation:

• Left side: \( \int e^{-y} \, dy = -e^{-y} + C \)
• Right side: \( \int e^{-x} \, dx = -e^{-x} + C \)

After integrating, we apply the initial conditions to solve for any constants of integration.
This step is crucial because it determines the specific solution to our initial value problem. In this particular example, the integration determined that \( C = 0 \), leading to the final solution where \( y = x \).
Thus, the act of integration bridges the gap from the separated equation to a concrete solution, fitting both the original differential equation and the given initial condition.