JEE Main 2024 (31 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

Q: How do I handle the absolute value signs in $\ln |\ln v| = \ln |y| + C$?

Since the problem explicitly states $x > 0$ and $y > 0$, we know that $v = x/y$ must also be positive. When we apply the initial condition $(e, 1)$, we find $C = 0$. This leaves us with $\ln |\ln v| = \ln |y|$. Because $y > 0$, we can drop the modulus on the right. For the left side, the condition $v=e$ makes $\ln v = 1$, which is positive, so the absolute value is naturally satisfied.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look closely at the structure of the differential equation. The term

(ln x - ln y)

is a massive hint. Recall the logarithmic property

ln a - ln b = ln (a / b)

. This suggests that the equation can be expressed entirely in terms of the ratio

x / y

, which is the hallmark of a homogeneous differential equation.

Hint 2

Hint 2 (The Arsenal): Since the equation is homogeneous, your primary tool is the substitution

v = x / y

, which implies

x = v y

. When you differentiate this with respect to

y

, remember to use the product rule:

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

. This transformation will allow you to separate the variables

v

and

y

Hint 3

Hint 3 (The Execution): After substituting and simplifying, you will arrive at the separable equation

y \frac{d v}{d y} = v ln v

. Your starting point for integration is

\int \frac{d v}{v l n v} = \int \frac{d y}{y}

. Use the substitution

u = ln v

to solve the left-hand integral.

Visualized Solution

Analyze the Equation $y \frac{d x}{d y} = x (ln x - ln y + 1)$

Given: $y \frac{d x}{d y} = x (ln x - ln y + 1)$
Initial Condition: Curve passes through $(e, 1)$
Constraints: $x > 0, y > 0$

Rearrange and Use Log Property

Divide by $y$ : $\frac{d x}{d y} = \frac{x}{y} (ln x - ln y + 1)$
Apply $ln a - ln b = ln \frac{a}{b}$ :
$\frac{d x}{d y} = \frac{x}{y} (ln \frac{x}{y} + 1)$

Substitution $x = v y$

Let $\frac{x}{y} = v \Rightarrow x = v y$
Differentiate both sides with respect to $y$

Differentiate using Product Rule

Using Product Rule: $\frac{d x}{d y} = v \cdot \frac{d}{d y} (y) + y \cdot \frac{d v}{d y}$
$\frac{d x}{d y} = v + y \frac{d v}{d y}$

Substitute into the Differential Equation

Substitute in $\frac{d x}{d y} = \frac{x}{y} (ln \frac{x}{y} + 1)$ :
$v + y \frac{d v}{d y} = v (ln v + 1)$

Simplify the Equation

Expand RHS: $v + y \frac{d v}{d y} = v ln v + v$
Subtract $v$ from both sides:
$y \frac{d v}{d y} = v ln v$

Separate the Variables

Rearrange terms: $\frac{d v}{v l n v} = \frac{d y}{y}$
Integrate both sides: $\int \frac{d v}{v l n v} = \int \frac{d y}{y}$

Integrate Both Sides

For $\int \frac{d v}{v l n v}$ , let $ln v = t \Rightarrow \frac{1}{v} d v = d t$
$\int \frac{d t}{t} = ln ∣ t ∣ = ln ∣ ln v ∣$

General Solution

$\int \frac{d y}{y} = ln ∣ y ∣ + C$
General Solution: $ln ∣ ln v ∣ = ln ∣ y ∣ + C$

Apply Initial Condition $(e, 1)$

At $(e, 1)$ : $x = e, y = 1$
Calculate $v$ : $v = \frac{x}{y} = \frac{e}{1} = e$
Substitute into general solution:
$ln ∣ ln e ∣ = ln ∣1∣ + C$

Solve for Constant $C$

$ln ∣1∣ = 0 + C$
$0 = 0 + C \Rightarrow C = 0$

Final Equation in $v$ and $y$

Substitute $C = 0$ : $ln ∣ ln v ∣ = ln ∣ y ∣$
Take antilog: $∣ ln v ∣ = ∣ y ∣$
Since $y > 0$ : $∣ ln v ∣ = y$

Final Form $∣ ln \frac{x}{y} ∣ = y$

Substitute $v = \frac{x}{y}$ back:
$∣ ln \frac{x}{y} ∣ = y$
This matches Option 3.

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

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

. At first glance, it appears complex, but we can simplify the logarithmic terms using the property

ln a - ln b = ln (\frac{a}{b})

By rewriting the equation as

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

, we reveal the structure of a homogeneous differential equation. The right-hand side depends solely on the ratio

\frac{x}{y}

, which is the hallmark of this class of problems.

The Power of Substitution

To solve this, we use the substitution

v = \frac{x}{y}

, which implies

x = v y

. Differentiating both sides with respect to

y

using the product rule, we obtain:

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

Substituting this into our original equation, we get

v + y \frac{d v}{d y} = v (ln v + 1)

. Expanding the right side yields

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

The

v

terms on both sides cancel out, leaving us with the simplified separable equation:

y \frac{d v}{d y} = v ln v

The Final Integration

We now separate the variables

v

and

y

to prepare for integration:

\frac{d v}{v ln v} = \frac{d y}{y}

To integrate the left side, we use the substitution

u = ln v

, which implies

d u = \frac{1}{v} d v

. The integral becomes

\int \frac{d u}{u} = ln ∣ u ∣ = ln ∣ ln v ∣

Integrating the right side gives

ln ∣ y ∣

. Combining these results, we obtain the general solution:

ln ∣ ln v ∣ = ln ∣ y ∣ + C

The Climax

Finding the Curve

We are given the initial condition that the curve passes through

(e, 1)

. At this point,

x = e

and

y = 1

, so the ratio

v = \frac{e}{1} = e

Plugging these values into our general solution:

ln ∣ ln e ∣ = ln ∣1∣ + C

. Since

ln e = 1

and

ln 1 = 0

, we find

ln ∣1∣ = 0 + C

, which simplifies to

0 = 0 + C

. Thus, $C = 0$ .

The equation becomes

ln ∣ ln v ∣ = ln ∣ y ∣

. Taking the antilog of both sides, we get

∣ ln v ∣ = ∣ y ∣

. Substituting

v = \frac{x}{y}

back into the expression, we arrive at the final result:

ln (\frac{x}{y}) = y

Common Traps & FAQs

Wait, why do we use the substitution

x = v y

instead of

y = v x

Great question! We look at the differential equation itself. It is given as

\frac{d x}{d y} = \dots

. This tells us that

x

is the dependent variable and

y

is the independent variable. Therefore, we express

x

as a function of

y

, making

x = v y

the correct substitution to simplify the expression into a form where variables can be separated.

Why did the

v

terms cancel out so perfectly in the middle of the problem?

That is the beauty of homogeneous differential equations! When you substitute

x = v y

into a truly homogeneous equation, the

v

terms are designed to simplify. If they didn't cancel or simplify into a form involving only

v

and

y

, it would be a sign that either the equation isn't homogeneous or a calculation error occurred. It is a built-in check for your work.

How do I handle the absolute value signs in

ln ∣ ln v ∣ = ln ∣ y ∣ + C

Since the problem explicitly states

x > 0

and

y > 0

, we know that

v = x / y

must also be positive. When we apply the initial condition

(e, 1)

, we find

C = 0

. This leaves us with

ln ∣ ln v ∣ = ln ∣ y ∣

. Because

y > 0

, we can drop the modulus on the right. For the left side, the condition

v = e

makes

ln v = 1

, which is positive, so the absolute value is naturally satisfied.

Animated Solution for Mathematics - Differential Equations: The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

Select Answer:

Visualized Solution

Analyze the Equation $y \frac{d x}{d y} = x (ln x - ln y + 1)$

Rearrange and Use Log Property

Substitution $x = v y$

Differentiate using Product Rule

Substitute into the Differential Equation

Simplify the Equation

Separate the Variables

Integrate Both Sides

General Solution

Apply Initial Condition $(e, 1)$

Solve for Constant $C$

Final Equation in $v$ and $y$

Final Form $∣ ln \frac{x}{y} ∣ = y$

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Power of Substitution

The Final Integration

The Climax

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ydydx​=x(lnx−lny+1)

Rearrange and Use Log Property

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

Differentiate using Product Rule

Substitute into the Differential Equation

Simplify the Equation

Separate the Variables

Integrate Both Sides

General Solution

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (e,1)

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

Final Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Final Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣lnyx​∣=y

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Power of Substitution

The Final Integration

The Climax

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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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