Let y(x) be the solution of the differential equation 2x² dy + (e^y - 2x) dx = 0, x 0. If y(e) = 1

Q: Wait, why can't we just use variable separation here? It looks like a simple equation.

That is a common trap! Variable separation works when you can write the equation as $f(y)dy = g(x)dx$. Here, the $e^y$ and $x$ terms are 'entangled' in a way that makes isolation impossible. We need a more powerful tool—the Bernoulli substitution—to untangle them.

Q: Why did we multiply by $e^{-y}$ specifically? How was I supposed to know that?

Great question. In Bernoulli equations of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, we divide by the non-linear term. Here, our 'non-linear' part is $e^y$. By multiplying by $e^{-y}$, we are essentially forcing the derivative of our substitution to appear in the equation. It is a strategic move to simplify the structure, not a random guess!

Q: I got stuck at the integrating factor. Is there a shortcut?

There is no shortcut for the integrating factor, but there is a 'sanity check.' Remember that the integrating factor $IF$ must always be a function of $x$ only. If you find your $IF$ contains $y$, you have made a mistake in your substitution or your identification of $P(x)$. Trust the process: $IF = e^{\int P(x) dx}$ is your best friend.

Question

Let $y(x)$ be the solution of the differential equation $2x^2 dy + (e^y - 2x) dx = 0, x > 0$. If $y(e) = 1$, then $y(1)$ is equal to :

Options:

(A) $0$

(B) $2$

(C) $\log_e 2$

(D) $\log_e(2e)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a differential equation that, at first glance, might seem like a chaotic mess of variables.

You see the equation $2x^2 dy + (e^y - 2x) dx = 0$ and your brain might immediately panic. In the world of differential equations, the first step is always about perception—learning to see the structure hidden beneath the surface.

### Phase 1: The Rearrangement

Our first goal is to bring this equation into a form that speaks to us. We want to isolate the derivative, $\frac{dy}{dx}$. By moving the $dx$ term to the right and dividing, we get:

$$\frac{dy}{dx} = -\frac{e^y - 2x}{2x^2}$$

Now, let's distribute that negative sign and split the fraction. We rewrite it as:

$$\frac{dy}{dx} = \frac{2x - e^y}{2x^2} = \frac{2x}{2x^2} - \frac{e^y}{2x^2}$$

Simplifying this, we get $\frac{dy}{dx} = \frac{1}{x} - \frac{e^y}{2x^2}$. If we move the $\frac{1}{x}$ term to the left, we have:

$$\frac{dy}{dx} - \frac{1}{x} = -\frac{e^y}{2x^2}$$

Look closely. If that $e^y$ term weren't there, this would be a simple linear differential equation. This is a **Bernoulli equation**, and we have a specific strategy to defeat it.

### Phase 2: The Bernoulli Transformation

To eliminate that $e^y$, we need to perform a surgical strike. We multiply the entire equation by $e^{-y}$:

$$e^{-y} \frac{dy}{dx} - \frac{1}{x} e^{-y} = -\frac{1}{2x^2}$$

Now, we introduce our hero: the substitution $z = e^{-y}$. If we differentiate this with respect to $x$, we get $\frac{dz}{dx} = -e^{-y} \frac{dy}{dx}$.

Substituting this into our equation, we get:

$$-\frac{dz}{dx} - \frac{z}{x} = -\frac{1}{2x^2}$$

Multiply by $-1$ to clean it up, and we arrive at the standard linear form:

$$\frac{dz}{dx} + \frac{z}{x} = \frac{1}{2x^2}$$

### Phase 3: The Integrating Factor

We have successfully transformed a non-linear equation into a standard linear equation. Here, $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{2x^2}$.

To solve this, we need the Integrating Factor ($IF$):

$$IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$$

With our $IF = x$, the solution is straightforward. We multiply our linear equation by $x$ and integrate:

$$z \cdot x = \int \left( \frac{1}{2x^2} \cdot x \right) dx = \int \frac{1}{2x} dx = \frac{1}{2} \ln x + C$$

### Phase 4: The Final Reveal

We substitute $z = e^{-y}$ back into our equation:

$$x e^{-y} = \frac{1}{2} \ln x + C$$

We are given the condition $y(e) = 1$. Let's plug in $x = e$ and $y = 1$ to find our constant $C$:

$$e \cdot e^{-1} = \frac{1}{2} \ln e + C \implies 1 = \frac{1}{2}(1) + C \implies C = \frac{1}{2}$$

Our particular solution is $x e^{-y} = \frac{1}{2} \ln x + \frac{1}{2}$. Finally, to find $y(1)$, we set $x = 1$:

$$1 \cdot e^{-y} = \frac{1}{2} \ln 1 + \frac{1}{2} \implies e^{-y} = \frac{1}{2} \implies e^y = 2$$

Taking the natural logarithm, we find the final answer:

**$y = \ln 2$**

Animated Solution for Mathematics - Differential Equations: Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange to Standard Form

Prepare for Substitution

Apply Substitution

Transform to Linear Form

Identify $P (x)$ and $Q (x)$

Calculate Integrating Factor

Write the General Solution

Resubstitute $z$

Apply Boundary Condition

Final Particular Solution

Calculate $y (1)$

Solve for $y$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=0,x>0. If y(e)=1, then y(1) is equal to :