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? Can't we just solve it as is?

Great question! If you keep it as $\frac{dy}{dx} = \frac{1}{y^2 - x}$, you have a non-linear equation that doesn't fit standard forms like Bernoulli or exact equations easily. By flipping it to $\frac{dx}{dy} = y^2 - x$, it transforms into a linear differential equation $\frac{dx}{dy} + x = y^2$, which is much more manageable. Always look for the path of least resistance in JEE problems!

Q: I got stuck on the integration by parts. Is there a shortcut?

For integrals of the form $\int P(y) e^y dy$ where $P(y)$ is a polynomial, you can use the tabular method (or DI method) to save time. Differentiate $y^2$ until it becomes zero and integrate $e^y$ repeatedly. It's a lifesaver in JEE exams and helps avoid sign errors!

Q: Why did we set $y=0$ to find the $x$-intercept?

Remember the coordinate geometry basics! The $x$-axis is defined by the equation $y=0$. Any point lying on the $x$-axis must have a $y$-coordinate of zero. Don't confuse this with finding the $y$-intercept, where you would set $x=0$. Always keep your axes definitions clear!

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

(C) $2 - e$

(D) $-e$

Sigma Percentile · Accepted Answer

### The Art of the Flip: Mastering Non-Linear Differential Equations

Differential equations are often like locked doors in a labyrinth. Sometimes, the key is right in front of you, but you are looking at the door from the wrong angle.

In this problem, we are given the equation $(y^2 - x) \frac{dy}{dx} = 1$. If you try to force this into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, you will quickly find yourself in a mathematical dead-end.

The equation is non-linear in $y$, and there is no obvious way to proceed.

### Phase 1: The Strategic Pivot

This is where the "JEE mindset" comes into play. When a differential equation looks like a tangled knot, stop and ask: *What if I change my perspective?*

Instead of treating $y$ as the dependent variable, let us treat $x$ as the dependent variable and $y$ as the independent one. By flipping the derivative, we get:

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

Suddenly, the equation breathes. Rearranging this gives us:

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

This is a classic Linear Differential Equation (LDE) in $x$!

### Phase 2: The Integrating Factor

Now that we have the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = 1$ and $Q(y) = y^2$, the path forward is illuminated.

We need the Integrating Factor ($I.F.$), which acts as a multiplier to make the left side of our equation a perfect derivative. The formula is:

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

Since $P(y) = 1$, our $I.F.$ is simply:

$$I.F. = e^{\int 1 dy} = e^y$$

It is elegant, simple, and powerful.

### Phase 3: The Integration Challenge

With our $I.F.$ in hand, the general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. Substituting our values, we get:

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

Now, we face the integral $\int y^2 e^y dy$. This requires Integration by Parts.

Applying the formula $\int u dv = uv - \int v du$ with $u = y^2$ and $dv = e^y dy$, we get $y^2 e^y - \int 2y e^y dy$.

We must apply the rule again to the remaining integral $\int 2y e^y dy$, which yields $2y e^y - 2e^y$. Combining these, we find the integral is $(y^2 - 2y + 2)e^y$.

### Phase 4: The Anchor of Initial Conditions

We now have the general solution:

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

But we are not done. We have a specific curve that satisfies $y(0) = 1$. This means when $x = 0$, $y = 1$.

Plugging these into our equation:

$$0 = (1)^2 - 2(1) + 2 + C e^{-1}$$

Simplifying this gives $0 = 1 + \frac{C}{e}$, which leads us directly to $C = -e$. Our particular solution is now locked in:

$$x = y^2 - 2y + 2 - e^{1-y}$$

### Phase 5: The Final Intercept

Finally, the problem asks for the abscissa of the point where the curve intersects the $x$-axis. As we know, the $x$-axis is the set of all points where $y = 0$.

Substituting $y = 0$ into our particular solution, we get:

$$x = (0)^2 - 2(0) + 2 - e^{1-0}$$

This simplifies to **$x = 2 - e$**.

And there it is! The curve intersects the $x$-axis at **$2 - e$**. You have successfully navigated the labyrinth. Keep this "flip" strategy in your toolkit—it is a powerful weapon for your JEE Advanced journey.

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

Analyzing the Differential Equation

Rearranging to Standard LDE Form

Calculating the Integrating Factor

Setting up the General Solution

Integration by Parts - First Application

Integration by Parts - Second Application

The General Solution for $x$

Applying the Initial Condition

Evaluating the Constant $C$

The Particular Solution

Finding the $x$ -intercept

Final Calculation

The Sigma Insight: Linear Differential Equations

The Art of the Flip

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

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