Let y = y(x) be the solution curve of the differential equation, (y² - x) dy/dx = 1, satisfying y(0)

Q: Wait, why do we flip the derivative to $\frac{dx}{dy}$? Isn't that illegal?

It is not illegal at all! In differential equations, we are essentially looking for a relationship between $x$ and $y$. If the equation is not easily solvable as $y=f(x)$, we can absolutely treat $x$ as a function of $y$. Flipping the derivative is a powerful technique when the equation is linear in $x$ but not in $y$.

Q: How do I know when to use the Integrating Factor method?

Whenever you can rearrange your first-order differential equation into the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P$ and $Q$ are functions of $y$ only, the Integrating Factor method is your best friend. It transforms a complex equation into a simple product rule derivative.

Q: Why does the constant $C$ matter so much in the final step?

The general solution $x = y^2 - 2y + 2 + C e^{-y}$ represents a whole family of curves. The initial condition $y(0) = 1$ acts like a GPS coordinate, pinning down the exact curve that passes through that specific point. Without finding $C$, you have the entire family, but not the specific path you need.

Question

Let $y = y(x)$ be the solution curve of the differential equation, $(y^2 - x) \frac{dy}{dx} = 1$, satisfying $y(0) = 1$. This curve intersects the $x$-axis at a point whose abscissa is:

Options:

(A) $2 - e$

(B) $2 + e$

(C) $2$

(D) $e$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the middle of a JEE Advanced exam, staring at the differential equation $(y^2 - x) \frac{dy}{dx} = 1$. At first glance, it looks like a nightmare.

You might try to separate the variables, but the $y^2$ term is locked in a struggle with the $x$ term. You might try to see if it is homogeneous, but it is not.

This is the moment where most students panic. But you are going to take a breath, look at the structure, and realize that sometimes, the best way to move forward is to change your perspective.

### The Reciprocal Trick

Look at the equation again: $(y^2 - x) \frac{dy}{dx} = 1$. If we try to solve for $\frac{dy}{dx}$, we get:

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

This is messy. But what if we flip the script? What if we treat $x$ as the dependent variable and $y$ as the independent variable?

By taking the reciprocal, we get $\frac{dx}{dy} = y^2 - x$. Suddenly, the equation transforms. By moving the $x$ term to the left, we get:

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

This is a classic linear differential equation in the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = 1$ and $Q(y) = y^2$. The panic fades, and the path becomes clear.

### The Integrating Factor

Now that we have our linear form, we need our secret weapon: the Integrating Factor (I.F.). The formula is:

$$I.F. = e^{\int P(y) dy}$$

Since $P(y) = 1$, our I.F. is simply $e^{\int 1 dy} = e^y$. This is the magic multiplier.

When we multiply our entire equation by $e^y$, the left side becomes the derivative of the product of $x$ and the I.F. That is:

$$\frac{d}{dy}(x e^y) = y^2 e^y$$

This is the beauty of the I.F. method; it turns a differential equation into a simple integration problem.

### The Dance of Integration

Now we integrate both sides:

$$x e^y = \int y^2 e^y dy + C$$

The integral $\int y^2 e^y dy$ is a classic case for Integration by Parts. We use the ILATE rule, choosing $u = y^2$ and $dv = e^y dy$.

The first pass gives us $y^2 e^y - \int 2y e^y dy$. We are not done yet! We apply Integration by Parts again to $\int 2y e^y dy$, which yields $2y e^y - 2e^y$.

Combining these, we get $x e^y = y^2 e^y - 2y e^y + 2e^y + C$. Dividing by $e^y$, we find the general solution:

$$x = y^2 - 2y + 2 + C e^{-y}$$

### The Final Reveal

We are almost at the finish line. We have the general solution, but we need the specific curve that satisfies $y(0) = 1$.

Substituting $x=0$ and $y=1$ into our equation, we get $0 = 1^2 - 2(1) + 2 + C e^{-1}$, which simplifies to $0 = 1 + C e^{-1}$. This tells us that $C = -e$.

Our specific solution is $x = y^2 - 2y + 2 - e^{1-y}$. The question asks for the $x$-intercept, which occurs when $y=0$.

Plugging $y=0$ into our equation, we get $x = 0^2 - 2(0) + 2 - e^{1-0}$, which simplifies to **$x = 2 - e$**.

And there it is! The abscissa of the intersection point is **$2 - e$**. You have navigated the complexity, applied the right tools, and arrived at the truth.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Inverting the Derivative

Identifying the Linear Form

Calculating the Integrating Factor

Setting up the General Solution

First Integration by Parts

Second Integration by Parts

Combining the Results

Applying Initial Condition

Finding the Constant C

Finding the X-axis Intersection

Final Calculation

Conclusion & Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Reciprocal Trick

The Integrating Factor

The Dance of Integration

The Final Reveal

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