JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . then $∣ (y (2))^{3} - 12 y (2) ∣$ is equal to:

Q: Why do we take the absolute value at the end?

The question specifically asks for $|(y(2))^3 - 12y(2)|$. The absolute value ensures that the final result is a non-negative magnitude, regardless of whether the expression inside evaluates to a positive or negative number. It is a common way to ensure a unique, well-defined answer in competitive exams.

Select Answer:

Progressive Hints

Hint 1

The Spark: Observe the differential equation

(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0

. Notice that every term, when expanded, has a total degree of 3. This is a classic indicator of a homogeneous differential equation. Your first task is to rearrange the terms to isolate

\frac{d y}{d x}

Hint 2

The Arsenal: Once you have

\frac{d y}{d x}

, the standard toolkit for homogeneous equations is the substitution

y = v x

. Remember to differentiate this with respect to

x

using the product rule:

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

. This will allow you to transform the equation into a separable form.

Hint 3

The Execution: After substituting and simplifying, you will arrive at the integral

\int \frac{2 ( 1 - v ^{2} )}{3 v - v ^{3}} d v = \int \frac{d x}{x}

. To solve the left side, consider the substitution

u = 3 v - v^{3}

. Notice how the numerator

2 (1 - v^{2}) d v

relates to the derivative of

u

Visualized Solution

Identify the Differential Equation

Given DE: $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$
Observe that every term is of degree 3, making it a Homogeneous Differential Equation.
Rearranging to find $\frac{d y}{d x}$ :
$\frac{d y}{d x} = \frac{y ( 5 x ^{2} - 3 y ^{2} )}{2 x ( x ^{2} - y ^{2} )}$

Substitution $y = v x$

Let $y = v x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d y}{d x} = v + x \frac{d v}{d x}$

Transforming the Equation

Substitute $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ into the DE:
$v + x \frac{d v}{d x} = \frac{v x ( 5 x ^{2} - 3 ( v x ) ^{2} )}{2 x ( x ^{2} - ( v x ) ^{2} )}$
Simplifying the right side by cancelling $x^{3}$ :
$v + x \frac{d v}{d x} = \frac{v ( 5 - 3 v ^{2} )}{2 ( 1 - v ^{2} )}$

Simplifying for $x \frac{d v}{d x}$

$x \frac{d v}{d x} = \frac{5 v - 3 v ^{3}}{2 ( 1 - v ^{2} )} - v$
$x \frac{d v}{d x} = \frac{5 v - 3 v ^{3} - 2 v ( 1 - v ^{2} )}{2 ( 1 - v ^{2} )}$
$x \frac{d v}{d x} = \frac{3 v - v ^{3}}{2 ( 1 - v ^{2} )}$

Variable Separation

Separating the variables $v$ and $x$ :
$\frac{2 ( 1 - v ^{2} )}{3 v - v ^{3}} d v = \frac{d x}{x}$
Integrating both sides:
$\int \frac{2 ( 1 - v ^{2} )}{v ( 3 - v ^{2} )} d v = \int \frac{d x}{x}$

Integration using Substitution

Let $u = 3 v - v^{3} ⟹ d u = (3 - 3 v^{2}) d v = 3 (1 - v^{2}) d v$
The integral becomes: $\frac{2}{3} \int \frac{d u}{u} = \int \frac{d x}{x}$
$\frac{2}{3} ln ∣3 v - v^{3} ∣ = ln ∣ x ∣ + C$

Simplifying the Logarithmic Form

$ln ∣3 v - v^{3} ∣^{\frac{2}{3}} = ln ∣ x ∣ + ln ∣ k ∣$
Taking antilog on both sides:
$(3 v - v^{3})^{\frac{2}{3}} = k x$
Raising both sides to the power of $\frac{3}{2}$ :
$3 v - v^{3} = C x^{\frac{3}{2}}$ (where $C = k^{\frac{3}{2}}$ )

Back-substitution of $v$

Substitute $v = \frac{y}{x}$ back into the equation:
$3 (\frac{y}{x}) - (\frac{y}{x})^{3} = C x^{\frac{3}{2}}$
$\frac{3 x ^{2} y - y ^{3}}{x ^{3}} = C x^{\frac{3}{2}}$
$3 x^{2} y - y^{3} = C x^{\frac{9}{2}}$

Finding the Constant $C$

Given $y (1) = 1$ , substitute $x = 1, y = 1$ :
$3 (1)^{2} (1) - (1)^{3} = C (1)^{\frac{9}{2}}$
$3 - 1 = C ⟹ C = 2$
The particular solution is: $3 x^{2} y - y^{3} = 2 x^{\frac{9}{2}}$

Evaluating at $x = 2$

Substitute $x = 2$ into the particular solution:
$3 (2)^{2} y (2) - (y (2))^{3} = 2 (2)^{\frac{9}{2}}$
$12 y (2) - (y (2))^{3} = 2 \cdot 2^{4} \cdot 2$
$12 y (2) - (y (2))^{3} = 322$

Final Answer

We need to find $∣ (y (2))^{3} - 12 y (2) ∣$ .
From the previous step: $12 y (2) - (y (2))^{3} = 322$
Multiplying by $- 1$ : $(y (2))^{3} - 12 y (2) = - 322$
Taking the absolute value:
$∣ (y (2))^{3} - 12 y (2) ∣ = ∣ - 322 ∣ = 322$

00:00 / 00:00

The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0

. By observing the degrees of the terms, we identify that every term has a total degree of three, confirming this is a homogeneous differential equation.

To begin, we rearrange the equation to isolate the slope,

\frac{d y}{d x}

\frac{d y}{d x} = \frac{y ( 5 x ^{2} - 3 y ^{2} )}{2 x ( x ^{2} - y ^{2} )}

The Master Key

To solve this, we employ the standard substitution for homogeneous equations:

y = v x

. Differentiating this with respect to

x

using the product rule yields:

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

Substituting these into our expression and simplifying by canceling the

x^{3}

terms, we obtain:

v + x \frac{d v}{d x} = \frac{v ( 5 - 3 v ^{2} )}{2 ( 1 - v ^{2} )}

The Integration Challenge

We now separate the variables by shifting the

v

term to the right side and finding a common denominator:

x \frac{d v}{d x} = \frac{5 v - 3 v ^{3} - 2 v + 2 v ^{3}}{2 ( 1 - v ^{2} )} = \frac{3 v - v ^{3}}{2 ( 1 - v ^{2} )}

Grouping the

v

terms with

d v

and the

x

terms with

d x

, we arrive at the following integral:

\int \frac{2 ( 1 - v ^{2} )}{3 v - v ^{3}} d v = \int \frac{d x}{x}

By letting

u = 3 v - v^{3}

, we find

d u = 3 (1 - v^{2}) d v

. Adjusting the constants, the integration results in:

\frac{2}{3} ln ∣3 v - v^{3} ∣ = ln ∣ x ∣ + C

The Final Stretch

Substituting

v = \frac{y}{x}

back into the equation, we get

\frac{2}{3} ln ∣ \frac{3 y}{x} - \frac{y ^{3}}{x ^{3}} ∣ = ln ∣ x ∣ + C

. Using the initial condition

y (1) = 1

, we find

C = 0

(since

ln ∣3 - 1∣ = 0

This simplifies to the relation:

(\frac{3 y}{x} - \frac{y ^{3}}{x ^{3}})^{2/3} = x \Rightarrow 3 x^{2} y - y^{3} = x^{3} \cdot x^{3/2} = x^{9/2}

Evaluating at

x = 2

, we have

3 (4) y - y^{3} = 2^{9/2}

, which simplifies to

12 y - y^{3} = 162

. The question asks for the absolute value

∣ y^{3} - 12 y ∣

Since

12 y - y^{3} = 162

, it follows that

y^{3} - 12 y = - 162

. Therefore, the final result is:

$∣ y^{3} - 12 y ∣ = 162$

Common Traps & FAQs

Wait, why do we use the substitution

y = v x

instead of something else?

Great question! In a homogeneous differential equation, the ratio of

y

x

is the key to unlocking the structure. By setting

y = v x

, we effectively 'scale' the equation, allowing the

x

terms to cancel out entirely. This reduces the complexity from a two-variable problem to a separable one-variable problem. It is the most elegant way to handle this specific type of symmetry.

Why can't we just use energy conservation or some other physics principle here?

While physics problems often involve energy conservation, this is a pure mathematical differential equation. We are not dealing with a physical system with conserved quantities, but rather a geometric relationship between

x

and

y

. Our goal is to find the family of curves that satisfy the given slope condition. Stick to the calculus tools—they are the right ones for this job!

Why do we take the absolute value at the end?

The question specifically asks for

∣ (y (2))^{3} - 12 y (2) ∣

. The absolute value ensures that the final result is a non-negative magnitude, regardless of whether the expression inside evaluates to a positive or negative number. It is a common way to ensure a unique, well-defined answer in competitive exams.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . then $∣ (y (2))^{3} - 12 y (2) ∣$ is equal to:

Select Answer:

Visualized Solution

Identify the Differential Equation

Substitution $y = v x$

Transforming the Equation

Simplifying for $x \frac{d v}{d x}$

Variable Separation

Integration using Substitution

Simplifying the Logarithmic Form

Back-substitution of $v$

Finding the Constant $C$

Evaluating at $x = 2$

Final Answer

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Master Key

The Integration Challenge

The Final Stretch

Similar Questions

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Select Answer:

Visualized Solution

Identify the Differential Equation

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=vx

Transforming the Equation

Simplifying for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xdxdv​

Variable Separation

Integration using Substitution

Simplifying the Logarithmic Form

Back-substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Final Answer

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Master Key

The Integration Challenge

The Final Stretch

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Simplifying for $x \frac{d v}{d x}$

Back-substitution of $v$

Finding the Constant $C$

Evaluating at $x = 2$

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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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 :

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