JEE Advanced 2016

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

Q: Why did we flip the derivative to $\frac{dx}{dy}$? Isn't $\frac{dy}{dx}$ the standard?

In JEE Advanced, the 'standard' is whatever makes the math easier! By flipping to $\frac{dx}{dy}$, we isolate the non-linear term on the right, turning a messy equation into a solvable Bernoulli form. Always look for the path of least resistance.

Q: How do I know when to use a Bernoulli substitution?

Look for the structure $\frac{dx}{dy} + P(y)x = Q(y)x^n$. If you see a term like $(x+2)^2$ on the right and $(x+2)$ on the left, that's your signal. It's a classic pattern designed to test your ability to recognize forms.

Q: Why does the intersection check with $y=(x+3)^2$ fail?

It's all about the range. When you substitute $y=(x+3)^2$ into the curve equation, the left side grows much faster than the right side can ever reach. The math proves they are in different 'worlds'—they simply never meet.

Select Answer:

* Multiple Correct

Visualized Solution

Analyzing the Differential Equation

Given: $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0$
Condition: $x > 0$
The curve passes through the point $(1, 3)$ .

Grouping Terms

Rearranging: $(x^{2} + 4 x + 4 + x y + 2 y) \frac{d y}{d x} = y^{2}$
Notice the perfect square: $x^{2} + 4 x + 4 = (x + 2)^{2}$
Factoring the rest: $x y + 2 y = y (x + 2)$

Simplifying the Bracket

Substituting the grouped terms:
$((x + 2)^{2} + y (x + 2)) \frac{d y}{d x} = y^{2}$
The term $(x + 2)$ is repeating, hinting at a substitution or specific form.

Flipping the Derivative

Expressing as $\frac{d x}{d y}$ :
$\frac{d x}{d y} = \frac{( x + 2 ) ^{2} + y ( x + 2 )}{y ^{2}}$
Separating the fraction:
$\frac{d x}{d y} = \frac{( x + 2 ) ^{2}}{y ^{2}} + \frac{x + 2}{y}$

Forming a Bernoulli Equation

Rearranging terms:
$\frac{d x}{d y} - \frac{1}{y} (x + 2) = \frac{1}{y ^{2}} (x + 2)^{2}$
This is a Bernoulli differential equation in terms of $(x + 2)$ .

Applying Substitution

Divide by $(x + 2)^{2}$ :
$(x + 2)^{- 2} \frac{d x}{d y} - \frac{1}{y} (x + 2)^{- 1} = \frac{1}{y ^{2}}$
Let $u = (x + 2)^{- 1} = \frac{1}{x + 2}$
Differentiating w.r.t $y$ : $\frac{d u}{d y} = - (x + 2)^{- 2} \frac{d x}{d y}$

Converting to Linear Form

Substituting into the equation:
$- \frac{d u}{d y} - \frac{1}{y} u = \frac{1}{y ^{2}}$
Multiplying by $- 1$ :
$\frac{d u}{d y} + \frac{1}{y} u = - \frac{1}{y ^{2}}$

Finding the Integrating Factor

Linear equation form: $\frac{d u}{d y} + P (y) u = Q (y)$
Here, $P (y) = \frac{1}{y}$ and $Q (y) = - \frac{1}{y ^{2}}$
Integrating Factor (I.F.) = $e^{\int P (y) d y} = e^{\int \frac{1}{y} d y}$
I.F. = $e^{l n y} = y$

General Solution

Solution formula: $u \cdot (I.F.) = \int Q (y) \cdot (I.F.) d y + C$
$u \cdot y = \int (- \frac{1}{y ^{2}}) \cdot y d y + C$
$u \cdot y = - \int \frac{1}{y} d y + C$
$u \cdot y = - ln y + C$

Applying the Initial Condition

Substitute back $u = \frac{1}{x + 2}$ :
$\frac{y}{x + 2} = - ln y + C$
The curve passes through $(1, 3)$ . Substitute $x = 1, y = 3$ :
$\frac{3}{1 + 2} = - ln 3 + C$
$1 = - ln 3 + C ⟹ C = 1 + ln 3$

Final Equation of the Curve

Substitute $C$ back into the equation:
$\frac{y}{x + 2} = - ln y + 1 + ln 3$
Using log properties: $ln 3 - ln y = ln (\frac{3}{y})$
Final Curve: $\frac{y}{x + 2} = 1 + ln (\frac{3}{y})$

Checking Option A & B

Let's find the intersection of the curve with the line $y = x + 2$ .
Substitute $y = x + 2$ into the curve equation:
$\frac{x + 2}{x + 2} = 1 + ln (\frac{3}{y})$
$1 = 1 + ln (\frac{3}{y}) ⟹ ln (\frac{3}{y}) = 0$

Finding the Exact Point

$ln (\frac{3}{y}) = 0 ⟹ \frac{3}{y} = e^{0} = 1$
$y = 3$
Since $y = x + 2$ , substituting $y = 3$ gives $3 = x + 2 ⟹ x = 1$ .
The only intersection point is $(1, 3)$ .

Checking Option C & D

Now check intersection with the parabola $y = (x + 3)^{2}$ .
Substitute into LHS of curve: $\frac{( x + 3 ) ^{2}}{x + 2} = \frac{x ^{2} + 6 x + 9}{x + 2} = x + 4 + \frac{1}{x + 2}$
Since $x > 0$ , LHS $> 4$ .

Comparing LHS and RHS

Substitute into RHS: $1 + ln (\frac{3}{( x + 3 ) ^{2}}) = 1 + ln 3 - 2 ln (x + 3)$
Since $x > 0$ , $x + 3 > 3 ⟹ ln (x + 3) > ln 3$
RHS $< 1 + ln 3 - 2 ln 3 = 1 - ln 3$
$1 - ln 3$ is negative (since $ln 3 \approx 1.098$ ).
LHS $> 4$ and RHS $< 0$ . They can never be equal!

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0

To simplify, we group the terms:

x^{2} + 4 x + 4

is the perfect square

(x + 2)^{2}

, and

x y + 2 y

factors to

y (x + 2)

. The equation becomes:

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

The Pivot

Changing Perspectives

Solving for

\frac{d y}{d x}

is inefficient. Instead, we invert the derivative to

\frac{d x}{d y}

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

Splitting the fraction yields:

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

Rearranging to isolate the linear term gives us a Bernoulli differential equation:

\frac{d x}{d y} - \frac{1}{y} (x + 2) = \frac{1}{y ^{2}} (x + 2)^{2}

The Transformation

Linearizing the Chaos

To linearize, we use the substitution

u = (x + 2)^{- 1}

. Differentiating with respect to

y

gives:

\frac{d u}{d y} = - (x + 2)^{- 2} \frac{d x}{d y}

Substituting this into our equation transforms the expression into a standard first-order linear differential equation:

\frac{d u}{d y} + \frac{1}{y} u = - \frac{1}{y ^{2}}

The Integrating Factor (IF) is calculated as:

I F = e^{\int \frac{1}{y} d y} = y

Multiplying the equation by the IF and integrating, we obtain:

u \cdot y = - ln y + C

Substituting

u = \frac{1}{x + 2}

back into the equation, we arrive at the general solution:

\frac{y}{x + 2} = - ln y + C

The Final Reveal

Intersection Analysis

Given the curve passes through

(1, 3)

, we substitute these values to find

C

\frac{3}{1 + 2} = - ln 3 + C \Rightarrow 1 = - ln 3 + C \Rightarrow C = 1 + ln 3

The final equation of the curve is:

\frac{y}{x + 2} = 1 + ln (\frac{3}{y})

Testing the intersection with the line

y = x + 2

1 = 1 + ln (\frac{3}{y}) \Rightarrow ln (\frac{3}{y}) = 0 \Rightarrow y = 3

Since

y = x + 2

, we find

x = 1

. The curve and the line intersect at exactly one point. Conversely, testing the parabola

y = (x + 3)^{2}

shows that the growth rates are incompatible, meaning they never intersect.

Common Traps & FAQs

Why did we flip the derivative to

\frac{d x}{d y}

? Isn't

\frac{d y}{d x}

the standard?

In JEE Advanced, the 'standard' is whatever makes the math easier! By flipping to

\frac{d x}{d y}

, we isolate the non-linear term on the right, turning a messy equation into a solvable Bernoulli form. Always look for the path of least resistance.

How do I know when to use a Bernoulli substitution?

Look for the structure

\frac{d x}{d y} + P (y) x = Q (y) x^{n}

. If you see a term like

(x + 2)^{2}

on the right and

(x + 2)

on the left, that's your signal. It's a classic pattern designed to test your ability to recognize forms.

Why does the intersection check with

y = (x + 3)^{2}

fail?

It's all about the range. When you substitute

y = (x + 3)^{2}

into the curve equation, the left side grows much faster than the right side can ever reach. The math proves they are in different 'worlds'—they simply never meet.

Animated Solution for Mathematics - Differential Equations: A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Grouping Terms

Simplifying the Bracket

Flipping the Derivative

Forming a Bernoulli Equation

Applying Substitution

Converting to Linear Form

Finding the Integrating Factor

General Solution

Applying the Initial Condition

Final Equation of the Curve

Checking Option A & B

Finding the Exact Point

Checking Option C & D

Comparing LHS and RHS

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Pivot

The Transformation

The Final Reveal

Similar Questions

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then, the number of points of intersection of $y = y_{1} (x)$ and $y = y_{2} (x)$ is

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

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

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

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

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

Let $y = y (x)$ be the solution of the differential equation $(x + 1) y^{'} - y = e^{3 x} (x + 1)^{2}$ , with $y (0) = \frac{1}{3}$ . Then, the point $x = - \frac{4}{3}$ for the curve $y = y (x)$ is:

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Grouping Terms

Simplifying the Bracket

Flipping the Derivative

Forming a Bernoulli Equation

Applying Substitution

Converting to Linear Form

Finding the Integrating Factor

General Solution

Applying the Initial Condition

Final Equation of the Curve

Checking Option A & B

Finding the Exact Point

Checking Option C & D

Comparing LHS and RHS

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Pivot

The Transformation

The Final Reveal

Similar Questions

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

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Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

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