Questions: Find the solution of the initial value problem dy/dx = (x-5)e^(-2y), y(5) = ln(5)^5

Transcript text: Find the solution of the initial value problem \[ \frac{d y}{d x}=(x-5) e^{-2 y}, \quad y(5)=\ln (5)^{5} \]

Solution

Solution Steps

To solve the initial value problem, we need to separate the variables and integrate both sides. Then, we apply the initial condition to find the constant of integration.

Step 1: Separate Variables and Integrate

Given the differential equation: \[ \frac{d y}{d x} = (x - 5) e^{-2 y} \] we separate the variables and integrate both sides: \[ \int e^{2 y} \, dy = \int (x - 5) \, dx \] This yields: \[ \frac{e^{2 y}}{2} = \frac{x^2}{2} - 5x + C \]

Step 2: Apply Initial Condition

We apply the initial condition \( y(5) = \ln(5)^5 \): \[ \frac{e^{2 \ln(5)^5}}{2} = \frac{5^2}{2} - 5 \cdot 5 + C \] Simplifying, we get: \[ \frac{e^{2 \ln(5)^5}}{2} = \frac{25}{2} - 25 + C \] \[ \frac{e^{2 \ln(5)^5}}{2} = -\frac{25}{2} + C \] Solving for \( C \): \[ C = \frac{25}{2} + \frac{e^{2 \ln(5)^5}}{2} \]

Step 3: Substitute Constant Back into General Solution

Substituting \( C \) back into the general solution: \[ \frac{e^{2 y}}{2} = \frac{x^2}{2} - 5x + \frac{25}{2} + \frac{e^{2 \ln(5)^5}}{2} \] Multiplying through by 2 to simplify: \[ e^{2 y} = x^2 - 10x + 25 + e^{2 \ln(5)^5} \]