JEE Main 2026 (22 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

Q: When I squared $\sqrt{9 + y^2} = 9 - y$, I got $y^2 + 9 = 81 + y^2 - 18y$. Did I do something wrong?

Not at all! You did exactly the right thing. The $y^2$ terms appear on both sides, which is a beautiful 'gift' from the math gods—they cancel out perfectly, turning a quadratic equation into a simple linear one. This is a sign that you are on the right track.

Q: Is it possible to solve this using polar coordinates instead of $y=vx$?

Absolutely! If you set $x = r \cos \theta$ and $y = r \sin \theta$, the term $\sqrt{x^2+y^2}$ becomes $r$. The differential $xdy - ydx$ transforms into $r^2 d\theta$. This is a very elegant way to solve it, though it requires a bit more comfort with trigonometric substitutions. Both methods lead to the same beautiful result.

Select Answer:

Visualized Solution

Analyze the Differential Equation

Given Equation: $x d y - y d x = x^{2} + y^{2} d x$
Initial Condition: $y (1) = 0$
Goal: Find $y (3)$

Rearranging to Standard Form

Divide by $d x$ : $x \frac{d y}{d x} - y = x^{2} + y^{2}$
Rearrange: $x \frac{d y}{d x} = y + x^{2} + y^{2}$

Substitution for Homogeneous DE

Let $y = v x$
Differentiating with respect to $x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting into the Equation

Substitute: $x (v + x \frac{d v}{d x}) = v x + x^{2} + (v x)^{2}$
Simplify RHS: $xv + x^{2} \frac{d v}{d x} = v x + x 1 + v^{2}$

Simplifying the Expression

Cancel $v x$ : $x^{2} \frac{d v}{d x} = x 1 + v^{2}$
Divide by $x$ : $x \frac{d v}{d x} = 1 + v^{2}$

Variable Separation

Separate variables: $\frac{d v}{1 + v ^{2}} = \frac{d x}{x}$

Integration Step

Integrate: $\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x}$
Result: $ln (v + 1 + v^{2}) = ln x + C$

Applying Initial Condition

At $x = 1, y = 0 ⟹ v = \frac{0}{1} = 0$
Substitute: $ln (0 + 1 + 0) = ln 1 + C$
$ln (1) = ln (1) + C ⟹ C = 0$

Finding the General Solution

$ln (v + 1 + v^{2}) = ln x$
$v + 1 + v^{2} = x$
Substitute $v = \frac{y}{x}$ : $\frac{y}{x} + 1 + \frac{y ^{2}}{x ^{2}} = x$
Multiply by $x$ : $y + x^{2} + y^{2} = x^{2}$

Solving for $y (3)$

Put $x = 3$ : $y + 3^{2} + y^{2} = 3^{2}$
$y + 9 + y^{2} = 9$
Rearrange: $9 + y^{2} = 9 - y$
Square both sides: $9 + y^{2} = (9 - y)^{2}$

Final Calculation

Expand: $9 + y^{2} = 81 + y^{2} - 18 y$
Cancel $y^{2}$ : $9 = 81 - 18 y$
Rearrange: $18 y = 72$
Final Answer: $y = 4$

Conclusion and Takeaway

Key Takeaway: For equations of form $M (x, y) d x + N (x, y) d y = 0$ where $M, N$ are homogeneous, use $y = v x$ .
The solution curve is a parabola: $y = \frac{x ^{2} - 1}{2}$
Final Value: $y (3) = 4$

00:00 / 00:00

The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the JEE journey. Today, we stand before a differential equation that might look like a tangled knot of variables, but it is a masterpiece of symmetry waiting to be unraveled.

We are looking at the equation

x d y - y d x = x^{2} + y^{2} d x

. At first glance, the presence of

x d y - y d x

and the radical

x^{2} + y^{2}

might feel intimidating.

However, observe the homogeneity; every term here has a degree of one. This is our signal, our 'green light' to deploy the powerful substitution

y = v x

The Transformation

To begin, we rearrange our equation into the standard form:

\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}

By setting

y = v x

, we invoke the product rule:

\frac{d y}{d x} = v + x \frac{d v}{d x}

. When we substitute this into our equation, the

v x

terms on both sides cancel out.

This leaves us with

x \frac{d v}{d x} = 1 + v^{2}

. This is the moment where the complexity collapses into a beautiful, separable form.

The Integration

Now, we separate our variables by bringing all the

v

terms to one side and the

x

terms to the other:

\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x}

This integral is a classic in the JEE repertoire. The left side integrates to

ln (v + 1 + v^{2})

, and the right side is simply

ln ∣ x ∣ + C

Applying our initial condition

y (1) = 0

is where we find our constant. Since

v = \frac{y}{x}

, at

x = 1

and

y = 0

, we find

v = 0

Substituting these into our integrated equation, we get

ln (0 + 1 + 0) = ln (1) + C

, which simplifies to

0 = 0 + C

. Thus, $C = 0$ .

The Final Reveal

The elegance of this result is breathtaking—the constant vanishes, leaving us with the relationship

ln (v + 1 + v^{2}) = ln x

. By exponentiating both sides, we strip away the logarithms to reveal:

v + 1 + v^{2} = x

Substituting

v = \frac{y}{x}

back into the equation, we get

\frac{y}{x} + 1 + \frac{y ^{2}}{x ^{2}} = x

. Multiplying through by

x

, we arrive at the final curve equation:

y + x^{2} + y^{2} = x^{2}

Now, we reach the final act: finding

y (3)

. Plugging

x = 3

into our equation, we get

y + 9 + y^{2} = 9

Rearranging this to

9 + y^{2} = 9 - y

and squaring both sides, the

y^{2}

terms vanish:

9 + y^{2} = 81 - 18 y + y^{2}

18 y = 72

Solving for

y

, we find the final answer: $y = 4$ .

Common Traps & FAQs

Why did we choose

y = v x

instead of

x = u y

? Is there a rule for this?

Great question! In this specific problem, the equation is easily rearranged into the form

\frac{d y}{d x} = f (\frac{y}{x})

. When you see terms like

x^{2} + y^{2}

and

x d y - y d x

, the substitution

y = v x

is the most natural path because it simplifies the differential term

x d y - y d x

into

x^{2} d v

. You could use

x = u y

, but it would likely lead to a more cumbersome algebraic path.

Wait, why does

ln (v + 1 + v^{2}) = ln x

imply

v + 1 + v^{2} = x

Because the natural logarithm function is a one-to-one (injective) function. If

ln (A) = ln (B)

, then

A

must equal

B

. By exponentiating both sides (applying

e^{x}

to both sides), we effectively 'remove' the log, leaving us with the algebraic relationship between

v

and

x

When I squared

9 + y^{2} = 9 - y

, I got

y^{2} + 9 = 81 + y^{2} - 18 y

. Did I do something wrong?

Not at all! You did exactly the right thing. The

y^{2}

terms appear on both sides, which is a beautiful 'gift' from the math gods—they cancel out perfectly, turning a quadratic equation into a simple linear one. This is a sign that you are on the right track.

Is it possible to solve this using polar coordinates instead of

y = v x

Absolutely! If you set

x = r cos θ

and

y = r sin θ

, the term

x^{2} + y^{2}

becomes

r

. The differential

x d y - y d x

transforms into

r^{2} d θ

. This is a very elegant way to solve it, though it requires a bit more comfort with trigonometric substitutions. Both methods lead to the same beautiful result.

Why do we ignore the negative root when solving

9 + y^{2} = 9 - y

When we square both sides of an equation, we sometimes introduce 'extraneous' solutions. After solving for

y

, it is always good practice to plug the value back into the original equation

9 + y^{2} = 9 - y

. Since

9 - y

must be positive for the square root to be defined, we ensure our result satisfies the original constraint.

Animated Solution for Mathematics - Differential Equations: Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Substitution for Homogeneous DE

Substituting into the Equation

Simplifying the Expression

Variable Separation

Integration Step

Applying Initial Condition

Finding the General Solution

Solving for $y (3)$

Final Calculation

Conclusion and Takeaway

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Integration

The Final Reveal

Similar Questions

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If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

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Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Substitution for Homogeneous DE

Substituting into the Equation

Simplifying the Expression

Variable Separation

Integration Step

Applying Initial Condition

Finding the General Solution

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3)

Final Calculation

Conclusion and Takeaway

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Integration

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve of the differential equation xdxdy​−y=y2+16x2​,y(1)=3 be y=y(x). Then y(2) is equal to :

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Solving for $y (3)$

Let the solution curve of the differential equation $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}, y (1) = 1$ . If for some $n \in N, y (2) \in [n - 1, n)$ , then $n$ is equal to

If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :