Question

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} = 1 + xe^{y-x}, -\sqrt{2}<x<\sqrt{2}, y(0) = 0$. Then, the minimum value of $y(x), x \in (-\sqrt{2}, \sqrt{2})$ is equal to :

• A. $(1 - \sqrt{3}) - \log_e(\sqrt{3} - 1)$
• B. $(1 + \sqrt{3}) - \log_e(\sqrt{3} - 1)$
• C. $(2 - \sqrt{3}) - \log_e 2$
• D. $(2 + \sqrt{3}) + \log_e 2$

Hint:

Whenever you see a functional form \(f(x, y)\) depending on \((y \pm x)\), substituting \(v = y \pm x\) is usually the fastest path to a separable equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a first-order differential equation.
The term \(e^{y-x}\) suggests a substitution \(v = y - x\) to reduce the equation to a separable form.
After solving for \(y(x)\), we use standard calculus techniques (setting the first derivative to zero) to find the minimum value.
Step 2: Key Formula or Approach:
1. Substitution: \(v = y - x \Rightarrow \frac{dv}{dx} = \frac{dy}{dx} - 1\).
2. Separation of variables: \(\int f(v) dv = \int g(x) dx\).
3. Condition for minimum: \(\frac{dy}{dx} = 0\).
Step 3: Detailed Explanation:
Let \(y - x = v \Rightarrow \frac{dy}{dx} = 1 + \frac{dv}{dx}\).
Substituting into the D.E.: \(1 + \frac{dv}{dx} = 1 + xe^v \Rightarrow \frac{dv}{dx} = xe^v\).
Rearranging for integration: \(e^{-v} dv = x dx\).
Integrating both sides: \(\int e^{-v} dv = \int x dx \Rightarrow -e^{-v} = \frac{x^2}{2} + C\).
Using initial condition \(y(0) = 0\): \(v(0) = 0 - 0 = 0\).
\(-e^0 = 0 + C \Rightarrow C = -1\).
So, \(-e^{-(y-x)} = \frac{x^2}{2} - 1 \Rightarrow e^{x-y} = 1 - \frac{x^2}{2}\).
Taking natural logarithm: \(x - y = \ln(1 - x^2/2) \Rightarrow y(x) = x - \ln(1 - x^2/2)\).
For minimum value, \(\frac{dy}{dx} = 0 \Rightarrow 1 + xe^{y-x} = 0 \Rightarrow xe^{y-x} = -1 \Rightarrow e^{x-y} = -x\).
Substituting this into our equation: \(-x = 1 - \frac{x^2}{2} \Rightarrow x^2 - 2x - 2 = 0\).
Solving the quadratic: \(x = \frac{2 \pm \sqrt{4 + 8}}{2} = 1 \pm \sqrt{3}\).
Given domain \(x \in (-\sqrt{2}, \sqrt{2})\), only \(x = 1 - \sqrt{3}\) is valid.
Substitute \(x = 1 - \sqrt{3}\) into \(y(x)\):
\(y_{min} = (1 - \sqrt{3}) - \ln(1 - \frac{(1 - \sqrt{3})^2}{2})\).
\(y_{min} = (1 - \sqrt{3}) - \ln(1 - \frac{4 - 2\sqrt{3}}{2}) = (1 - \sqrt{3}) - \ln(\sqrt{3} - 1)\).
Step 4: Final Answer:
The minimum value is \((1 - \sqrt{3}) - \log_e(\sqrt{3} - 1)\).