Let y = y(x) be the solution of the differential equation dy/dx = 1 + xe^(y-x), -√2

Q: Why did we choose $v = y - x$? Could we have used something else?

That is a brilliant question! In JEE Advanced, we look for patterns. The term $y-x$ appearing in the exponent is a 'signature' of a specific type of differential equation. If we had tried to solve it as a linear differential equation or by separating variables directly, we would have hit a wall. By setting $v = y - x$, we essentially 'linearize' the exponent, turning a complex transcendental equation into a simple variable-separable one. Always look for the 'troublemaker' term in the equation—that is usually where the substitution lies.

Q: Why do we reject $x = 1 + \sqrt{3}$? It is a valid root of the quadratic, right?

You are absolutely correct that $1 + \sqrt{3}$ is a root of the quadratic equation $x^2 - 2x - 2 = 0$. However, in mathematics, we must always respect the 'boundary conditions' or the domain of the problem. The problem explicitly states that $x$ must lie in the interval $(-\sqrt{2}, \sqrt{2})$. Since $\sqrt{3} \approx 1.732$ and $\sqrt{2} \approx 1.414$, the value $1 + 1.732 = 2.732$ is clearly outside the allowed domain. We must discard it because the function $y(x)$ is not defined or valid outside that interval.

Q: Is it always true that $\frac{dy}{dx} = 0$ at a minimum?

For a differentiable function on an open interval, yes! The condition $\frac{dy}{dx} = 0$ is the necessary condition for a local extremum (a minimum or a maximum). Since we are looking for the minimum value of $y(x)$, setting the derivative to zero allows us to pinpoint the exact $x$-coordinate where the curve 'turns around'. It is the most powerful tool in our calculus toolkit for optimization.

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 :

Options:

(A) $(2 - \sqrt{3}) - \log_e 2$

(B) $(2 + \sqrt{3}) + \log_e 2$

(C) $(1 + \sqrt{3}) - \log_e(\sqrt{3} - 1)$

(D) $(1 - \sqrt{3}) - \log_e(\sqrt{3} - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the differential equation:

$$\frac{dy}{dx} = 1 + xe^{y-x}$$

When you see an expression like $e^{y-x}$, your intuition should immediately scream "substitution!" The term $y-x$ is the "troublemaker" preventing us from separating the variables $x$ and $y$.

By defining a new variable $v = y - x$, we are essentially performing a coordinate transformation. Differentiating this with respect to $x$, we get:

$$\frac{dy}{dx} - 1 = \frac{dv}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = 1 + \frac{dv}{dx}$$

This is the key that unlocks the door.

### The Dance of Integration

Now, watch the magic happen. We substitute our new expression for $\frac{dy}{dx}$ back into the original equation:

$$1 + \frac{dv}{dx} = 1 + xe^v$$

The constants cancel out, leaving us with the beautiful, clean equation:

$$\frac{dv}{dx} = xe^v$$

This is a classic variable-separable differential equation. We move the $v$ terms to the left and the $x$ terms to the right:

$$e^{-v} dv = x dx$$

Integrating both sides, we obtain:

$$-e^{-v} = \frac{x^2}{2} + C$$

We have successfully tamed the beast.

### Finding the Constant and the Function

We are given the initial condition $y(0) = 0$. Since $v = y - x$, at $x=0$, $v$ is also $0$.

Plugging these into our integrated equation, we find:

$$-e^0 = 0 + C \quad \Rightarrow \quad C = -1$$

Our equation becomes:

$$-e^{-(y-x)} = \frac{x^2}{2} - 1$$

Rearranging this, we find $e^{x-y} = 1 - \frac{x^2}{2}$. Taking the natural logarithm, we finally isolate $y$:

$$y(x) = x - \ln\left(1 - \frac{x^2}{2}\right)$$

This is the explicit function we are working with.

### The Critical Moment

To find the minimum value, we need to find where the slope of the curve is zero. Setting $\frac{dy}{dx} = 0$ in our original equation $1 + xe^{y-x} = 0$, we get:

$$xe^{y-x} = -1$$

Using our previous relation $e^{y-x} = \frac{1}{e^{x-y}} = \frac{1}{1 - \frac{x^2}{2}} = \frac{2}{2 - x^2}$, we substitute this into our derivative condition:

$$1 + x\left(\frac{2}{2 - x^2}\right) = 0$$

This simplifies to the quadratic equation:

$$x^2 - 2x - 2 = 0$$

Solving this using the quadratic formula gives $x = 1 \pm \sqrt{3}$.

### The Final Verdict

We must be vigilant. The domain of the function is defined by $1 - \frac{x^2}{2} > 0$, which implies $x^2 < 2$, or $x \in (-\sqrt{2}, \sqrt{2})$.

The root $x = 1 + \sqrt{3} \approx 2.732$ is outside our domain. Thus, we accept $x = 1 - \sqrt{3}$.

Substituting this back into our function $y(x)$, we calculate the minimum value:

**$$y_{min} = (1 - \sqrt{3}) - \ln(\sqrt{3} - 1)$$**

You have navigated the complexity and arrived at the truth. Keep this confidence; it is the hallmark of a true mathematician.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 1 + x e^{y - x}, - 2 < x < 2, y (0) = 0$ , then, the minimum value of $y (x), x \in (- 2, 2)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Substitution Strategy: $y - x = v$

Transforming the Equation

Separating Variables and Integrating

Finding the Constant $C$

Expressing $y$ in terms of $x$

Condition for Minimum Value

Solving for Critical Point $x$

Finding the Valid Critical Point

Calculating the Minimum Value

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Dance of Integration

Finding the Constant and the Function

The Critical Moment

The Final Verdict

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Substitution Strategy: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y−x=v

Transforming the Equation

Separating Variables and Integrating

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Condition for Minimum Value

Solving for Critical Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x