The solution of the differential equation (1+y²) + (x - e^(tan⁻¹y))dy/dx = 0, is (a) xe^(2tan⁻¹y) =

Q: Wait, why are we allowed to just flip the equation to $\frac{dx}{dy}$? Isn't $\frac{dy}{dx}$ the standard way?

That is a brilliant question! In JEE, we often get stuck in the habit of always solving for $y$ in terms of $x$. But a differential equation is just a relationship between variables. If the equation is linear in $x$ but non-linear in $y$, treating $x$ as the function of $y$ is not just allowed—it is the most elegant path to the solution. It is like choosing the path of least resistance in a circuit!

Q: How do I know when to use the Integrating Factor method versus separation of variables?

Great intuition. Separation of variables works when you can isolate all $x$'s and $dx$'s on one side and all $y$'s and $dy$'s on the other. Here, the $x$ and $y$ terms are 'tangled' together in a way that makes separation impossible. Whenever you see a linear term of one variable added to a function of the other, the linear form $\frac{dx}{dy} + P(y)x = Q(y)$ is your best friend.

Q: I got stuck at the integral $\int \frac{e^{2\tan^{-1}y}}{1+y^2} dy$. How do I handle that $\tan^{-1}y$?

Don't let the inverse trig function intimidate you! Whenever you see $\tan^{-1}y$ alongside its derivative $\frac{1}{1+y^2}$ in an integral, it is a massive 'hint' from the examiner to use substitution. Let $u = \tan^{-1}y$. The term $\frac{1}{1+y^2} dy$ becomes $du$, and the integral transforms into the much simpler $\int e^{2u} du$.

Q: Why did we multiply by 2 at the end? Is it wrong if I leave the constant as $C$?

Mathematically, $C$ and $k$ are both arbitrary constants. However, in competitive exams, we often manipulate the constant to match the format of the given options. If the options don't have fractions, multiplying by a constant factor is a standard way to clean up the expression while keeping the physics/math integrity intact.

Q: Is it possible to solve this by keeping $\frac{dy}{dx}$?

Technically, you could try, but you would end up with a non-linear differential equation in $y$, which is significantly harder—perhaps even impossible—to solve using standard JEE methods. Always look for the 'linear' path first; it is the hallmark of a smart problem solver.

Question

The solution of the differential equation $(1+y^2) + (x - e^{	an^{-1}y})\frac{dy}{dx} = 0$, is

Options:

(A) $xe^{2	an^{-1}y} = e^{	an^{-1}y} + k$

(B) $(x-2) = ke^{2	an^{-1}y}$

(C) $2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$

(D) $xe^{	an^{-1}y} = 	an^{-1}y + k$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, tangled knot of variables. The equation $(1+y^2) + (x - e^{	an^{-1}y})\frac{dy}{dx} = 0$ looks intimidating at first glance.

It is a classic JEE Advanced trap—a problem designed to test not just your calculation speed, but your ability to choose the right perspective.

Most students will immediately try to isolate $\frac{dy}{dx}$ and attempt to solve it as a function of $x$. But look closer. Do you see how $y$ is trapped inside the transcendental $	an^{-1}y$ function, while $x$ stands alone, linear and exposed?

This is the problem whispering its secret to you: "Change your point of view."

### The Shift: Flipping the Lens

When we encounter a differential equation where one variable is "messy" and the other is "clean," we must be brave enough to flip the script. By treating $x$ as the dependent variable and $y$ as the independent variable, we transform the equation.

We multiply the entire expression by $\frac{dx}{dy}$, effectively turning the equation into $(1+y^2)\frac{dx}{dy} + (x - e^{	an^{-1}y}) = 0$.

Suddenly, the complexity begins to dissolve. By dividing through by $(1+y^2)$, we arrive at the standard linear form:

$$\frac{dx}{dy} + \left( \frac{1}{1+y^2} ight)x = \frac{e^{	an^{-1}y}}{1+y^2}$$

This is the beauty of the linear differential equation form $\frac{dx}{dy} + P(y)x = Q(y)$. We have successfully linearized the problem.

### The Integrating Factor: The Key to the Lock

Now that we have identified $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{e^{	an^{-1}y}}{1+y^2}$, we need our "skeleton key"—the Integrating Factor ($I.F.$).

The $I.F.$ is defined as $e^{\int P(y) dy}$. Since the integral of $\frac{1}{1+y^2}$ is simply $	an^{-1}y$, our $I.F.$ becomes $e^{	an^{-1}y}$.

Think of the $I.F.$ as a mathematical catalyst. When we multiply our entire equation by $e^{	an^{-1}y}$, the left-hand side collapses into the derivative of a product: $\frac{d}{dy}(x \cdot e^{	an^{-1}y})$.

This is the moment of elegance where the chaos of the original equation organizes itself into a perfect derivative.

### The Final Integration: Crossing the Finish Line

We are left with the integral:

$$x \cdot e^{	an^{-1}y} = \int \left( \frac{e^{	an^{-1}y}}{1+y^2} ight) \cdot e^{	an^{-1}y} dy + C$$

Simplifying the integrand gives us $\int \frac{e^{2	an^{-1}y}}{1+y^2} dy$. I know this looks like a daunting integral, but take a breath.

Let $u = 	an^{-1}y$. Then $du = \frac{1}{1+y^2} dy$. The integral transforms into $\int e^{2u} du$, which is simply $\frac{e^{2u}}{2}$.

Substituting back, we get $\frac{e^{2	an^{-1}y}}{2} + C$.

Finally, we multiply by 2 to clear the fraction, yielding the **final solution**:

$$2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$$

You have navigated the complexity, identified the hidden structure, and arrived at the solution with precision. This is not just math; this is the triumph of logical clarity over confusion.

Animated Solution for Mathematics - Differential Equations: The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

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Visualized Solution

Analyze the Equation

Rearranging Terms

Forming the Linear Equation

Identifying $P (y)$ and $Q (y)$

Calculating the Integrating Factor

Setting up the General Solution

Evaluating the Integral

Final Simplification

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Shift

The Integrating Factor

The Final Integration

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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; } The solution of the differential equation (1+y2)+(x−etan−1y)dxdy​=0, is