If the solution curve of the differential equation ((tan⁻¹ y) - x) dy = (1 + y²) dx passes through t

Q: Wait, why do we switch to $\frac{dx}{dy}$? Isn't $\frac{dy}{dx}$ the standard way?

That is a great observation! In many textbook problems, we are conditioned to solve for $\frac{dy}{dx}$. However, in JEE Advanced, the ability to pivot your perspective is crucial. If solving for $\frac{dy}{dx}$ leads to a non-linear, unsolvable mess, it is a massive hint that the equation is actually linear in $x$ when viewed as a function of $y$. Always look for the structure $\frac{dx}{dy} + P(y)x = Q(y)$ when $\frac{dy}{dx}$ fails you.

Q: How do I know it's a linear differential equation?

A linear differential equation in $x$ (where $x$ is the dependent variable and $y$ is the independent variable) will always have the form $\frac{dx}{dy} + P(y)x = Q(y)$. If you can rearrange your equation into this form, where $P(y)$ and $Q(y)$ are functions of $y$ only, you have a guaranteed path to the solution using the integrating factor method.

Q: Why is the integral of $t e^t$ so easy?

It looks intimidating at first, but once you use the substitution $t = \tan^{-1} y$, the term $\frac{1}{1+y^2} dy$ becomes $dt$. This transforms the integral into $\int t e^t dt$. This is a classic application of Integration by Parts. Remember the LIATE rule—algebraic functions come before exponential ones, so we set $u=t$ and $dv=e^t dt$. It simplifies beautifully to $(t-1)e^t$.

Question

If the solution curve of the differential equation $((	an^{-1} y) - x) dy = (1 + y^2) dx$ passes through the point $(1, 0)$ then the abscissa of the point on the curve whose ordinate is $	an(1)$ is :

Options:

(A) $2e$

(B) $\frac{2}{e}$

(C) 2

(D) $\frac{1}{e}$

Sigma Percentile · Accepted Answer

### The Art of the Perspective Shift

Welcome, fellow traveler on the JEE journey. Today, we are going to tackle a differential equation that, at first glance, might make your heart skip a beat. The equation is $((	an^{-1} y) - x) dy = (1 + y^2) dx$.

It looks like a tangled knot, doesn't it? If you try to force it into the standard $\frac{dy}{dx}$ form, you will find yourself staring at a non-linear nightmare involving $	an^{-1} y$.

But here is the secret: in mathematics, as in life, when the path ahead is blocked, sometimes the best solution is to change your perspective. Instead of asking how $y$ changes with $x$, let us ask how $x$ changes with $y$. By rearranging our equation to solve for $\frac{dx}{dy}$, we unlock a hidden, elegant structure.

### Unveiling the Linear Form

Let us perform the division by $(1+y^2) dy$. The equation transforms into:

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

With a little algebraic housekeeping, we can write this as:

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

Suddenly, the fog clears. This is a classic Linear Differential Equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$. Here, our $P(y)$ is $\frac{1}{1+y^2}$ and our $Q(y)$ is $\frac{	an^{-1} y}{1+y^2}$.

### The Magic of the Integrating Factor

Now, we reach for our most powerful tool: the Integrating Factor (I.F.). The formula is $I.F. = e^{\int P(y) dy}$. Substituting our $P(y)$, we get:

$$I.F. = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$$

This is the magic key that will allow us to collapse the left side of our differential equation into a single derivative. The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.

Plugging in our values, we get:

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

### The Elegance of Substitution

Do not let the right-hand integral intimidate you. Look closely at the integrand: $\frac{	an^{-1} y}{1 + y^2} e^{	an^{-1} y}$. Notice that $	an^{-1} y$ and its derivative $\frac{1}{1+y^2}$ are both present.

This is a signal to use substitution. Let $t = 	an^{-1} y$, which implies $dt = \frac{1}{1+y^2} dy$. The integral simplifies to $\int t e^t dt$.

Using Integration by Parts, where we treat $t$ as the first function and $e^t$ as the second, we get:

$$\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t$$

Substituting back, we find the integral is $(	an^{-1} y - 1) e^{	an^{-1} y}$.

### The Final Stretch

We are almost there! Our general solution is:

$$x \cdot e^{	an^{-1} y} = (	an^{-1} y - 1) e^{	an^{-1} y} + C$$

We are given that the curve passes through $(1, 0)$. Substituting $x=1$ and $y=0$, we get $1 \cdot e^0 = (0 - 1) e^0 + C$, which simplifies to $1 = -1 + C$, giving us $C = 2$.

The specific solution is:

$$x \cdot e^{	an^{-1} y} = (	an^{-1} y - 1) e^{	an^{-1} y} + 2$$

The question asks for the abscissa when $y = 	an(1)$. Since $	an^{-1}(	an(1)) = 1$, we substitute this into our equation:

$$x \cdot e^1 = (1 - 1) e^1 + 2$$

This simplifies to $x \cdot e = 2$, or **$x = \frac{2}{e}$**. You have navigated the complexity and arrived at the truth.

Animated Solution for Mathematics - Differential Equations: If the solution curve of the differential equation $((tan^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point $(1, 0)$ then the abscissa of the point on the curve whose ordinate is $tan (1)$ is :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange to Standard Form $\frac{d x}{d y} + P x = Q$

Identify $P (y)$ and $Q (y)$

Calculate the Integrating Factor (I.F.)

Set up the General Solution

Solve the Integral using Substitution

Integration by Parts

The Complete General Solution

Apply Initial Condition $(1, 0)$

Final Specific Solution

Find Abscissa for $y = tan (1)$

Final Calculation and Answer

The Sigma Insight: Linear Differential Equations

The Art of the Perspective Shift

Unveiling the Linear Form

The Magic of the Integrating Factor

The Elegance of Substitution

The Final Stretch

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearrange to Standard Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dydx​+Px=Q

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(y) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(y)