JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of $y (10)$ equals :

Q: Wait, why do we need the absolute value in the natural log? Can't we just ignore it?

That is a great question! The integral of $\frac{1}{u}$ is $\ln|u| + C$ because the natural logarithm is only defined for positive numbers. When you integrate, you don't know if the expression inside is positive or negative, so the absolute value ensures the math remains valid. However, you cannot just ignore it—you must use your initial condition to determine the sign of the expression inside the absolute value.

Q: Why is the sign negative when we remove the absolute value? How do I know?

This is the classic trap! Look at your initial condition: $y(4) = 1.5$. Plug this into the expression $\frac{y-3}{y}$. You get $\frac{1.5-3}{1.5} = \frac{-1.5}{1.5} = -1$. Since the problem states the slope is never zero, the curve cannot cross the lines $y=0$ or $y=3$. Because the expression is negative at $x=4$, it must remain negative for all $x > 2$. Therefore, when you remove the absolute value, you must include a negative sign.

Q: Is there a faster way to solve this without finding the constant $C$?

Yes, absolutely! You can use definite integration. Instead of finding $C$, you can integrate from the initial state $(x=4, y=1.5)$ to the final state $(x=10, y=y_{final})$. The integral on the left would be $\int_{1.5}^{y} \frac{dy}{y^2-3y}$ and the integral on the right would be $\int_{4}^{10} \frac{dx}{x^2-4}$. This approach automatically accounts for the constant and is often less prone to algebraic errors.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a separable differential equation. Your first goal is to isolate the variables

x

and

y

on opposite sides of the equation. Visualize the curve as a path in the

x y

-plane; the condition

y (4) = \frac{3}{2}

is your anchor point on this path. The trap lies in the absolute value that appears during integration—you must determine the sign of the expression

\frac{y - 3}{y}

using your anchor point.

Hint 2

Hint 2: The Arsenal. To integrate the separated variables, you need to use partial fraction decomposition. For the

y

-side, factor

y^{2} - 3 y

y (y - 3)

. For the

x

-side, factor

x^{2} - 4

(x - 2) (x + 2)

. Remember that

\int \frac{1}{u} d u = ln ∣ u ∣ + C

. The key is to keep the constant of integration

C

and solve for it using the initial condition

y (4) = \frac{3}{2}

Hint 3

Hint 3: The Execution. Start by writing the separated equation:

\frac{1}{3} \int (\frac{1}{y - 3} - \frac{1}{y}) d y = \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2}) d x

. After integrating, you will have

\frac{1}{3} ln ∣ \frac{y - 3}{y} ∣ = \frac{1}{4} ln ∣ \frac{x - 2}{x + 2} ∣ + C

. Use

x = 4, y = 1.5

to find

C

, then substitute

x = 10

to solve for

y

Visualized Solution

The Differential Equation

Given differential equation: $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0$
Initial condition: $y (4) = \frac{3}{2}$
Goal: Find the value of $y (10)$

Variable Separation

Rearrange terms to separate variables:
$(x^{2} - 4) d y = (y^{2} - 3 y) d x$
Divide to group $y$ with $d y$ and $x$ with $d x$ :
$\frac{d y}{y ^{2} - 3 y} = \frac{d x}{x ^{2} - 4}$

Partial Fraction Decomposition (LHS)

Factor the denominator on the Left Hand Side (LHS): $y^{2} - 3 y = y (y - 3)$
Apply partial fractions:
$\frac{1}{y ( y - 3 )} = \frac{1}{3} (\frac{1}{y - 3} - \frac{1}{y})$

Partial Fraction Decomposition (RHS)

Factor the denominator on the Right Hand Side (RHS): $x^{2} - 4 = (x - 2) (x + 2)$
Apply partial fractions:
$\frac{1}{x ^{2} - 4} = \frac{1}{4} (\frac{1}{x - 2} - \frac{1}{x + 2})$

Setting Up the Integrals

Substitute the decomposed fractions back into the equation:
$\frac{1}{3} \int (\frac{1}{y - 3} - \frac{1}{y}) d y = \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2}) d x$

Integration

Integrate each term using $\int \frac{1}{t} d t = ln ∣ t ∣$ :
$\frac{1}{3} (ln ∣ y - 3∣ - ln ∣ y ∣) = \frac{1}{4} (ln ∣ x - 2∣ - ln ∣ x + 2∣) + C$
Combine using log properties ( $ln a - ln b = ln \frac{a}{b}$ ):
$\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln \frac{x - 2}{x + 2} + C$

Finding the Constant $C$ (Setup)

Use the initial condition: at $x = 4$ , $y = \frac{3}{2} = 1.5$
Substitute these values into the integrated equation:
$\frac{1}{3} ln \frac{1.5 - 3}{1.5} = \frac{1}{4} ln \frac{4 - 2}{4 + 2} + C$

Calculating $C$

Simplify the LHS: $\frac{1}{3} ln \frac{- 1.5}{1.5} = \frac{1}{3} ln ∣ - 1∣ = \frac{1}{3} ln 1 = 0$
Simplify the RHS: $\frac{1}{4} ln \frac{2}{6} + C = \frac{1}{4} ln (\frac{1}{3}) + C$
Equate and solve for $C$ :
$0 = - \frac{1}{4} ln 3 + C \Rightarrow C = \frac{1}{4} ln 3$

Substituting $x = 10$

Substitute $C = \frac{1}{4} ln 3$ into the general equation.
We need to find $y$ when $x = 10$ . Substitute $x = 10$ :
$\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln \frac{10 - 2}{10 + 2} + \frac{1}{4} ln 3$

Simplifying the RHS

Simplify the fraction: $\frac{10 - 2}{10 + 2} = \frac{8}{12} = \frac{2}{3}$
The RHS becomes: $\frac{1}{4} ln (\frac{2}{3}) + \frac{1}{4} ln 3$
Factor out $\frac{1}{4}$ and use log addition ( $ln a + ln b = ln (ab)$ ):
$\frac{1}{4} (ln \frac{2}{3} + ln 3) = \frac{1}{4} ln (\frac{2}{3} \cdot 3) = \frac{1}{4} ln 2$

Isolating the Log Term

Equate LHS and the simplified RHS:
$\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln 2$
Multiply both sides by 3:
$ln \frac{y - 3}{y} = \frac{3}{4} ln 2$
Use the power rule for logs ( $k ln a = ln a^{k}$ ):
$ln \frac{y - 3}{y} = ln (2^{\frac{3}{4}}) = ln (8^{\frac{1}{4}})$

The Absolute Value Trap

Remove the natural log from both sides:
$\frac{y - 3}{y} = 8^{\frac{1}{4}}$
Check the sign of $\frac{y - 3}{y}$ using the initial condition $y (4) = 1.5$ :
$\frac{1.5 - 3}{1.5} = - 1 < 0$
Since the curve is continuous and slope is never zero, the expression remains negative.
Therefore, $\frac{y - 3}{y} = - 8^{\frac{1}{4}}$

Solving for $y (10)$

We have the equation: $\frac{y - 3}{y} = - 8^{\frac{1}{4}}$
Multiply by $y$ : $y - 3 = - y \cdot 8^{\frac{1}{4}}$
Bring all $y$ terms to one side: $y + y \cdot 8^{\frac{1}{4}} = 3$
Factor out $y$ : $y (1 + 8^{\frac{1}{4}}) = 3$
Final answer: $y = \frac{3}{1 + 8 ^{\frac{1}{4}}}$

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

We are given the differential equation

(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0

with the initial condition

y (4) = \frac{3}{2}

. Our objective is to determine the value of

y

when

x = 10

The Art of Separation

To solve this, we first separate the variables by moving the

d x

term to the right and grouping the

y

terms with

d y

and the

x

terms with

d x

. This yields the following symmetric form:

\frac{d y}{y ^{2} - 3 y} = \frac{d x}{x ^{2} - 4}

The Power of Decomposition

We simplify the denominators using partial fraction decomposition. For the left side, we factor

y^{2} - 3 y = y (y - 3)

, which allows us to write:

\frac{1}{y ( y - 3 )} = \frac{1}{3} (\frac{1}{y - 3} - \frac{1}{y})

For the right side, we use the difference of squares

x^{2} - 4 = (x - 2) (x + 2)

, leading to:

\frac{1}{x ^{2} - 4} = \frac{1}{4} (\frac{1}{x - 2} - \frac{1}{x + 2})

Substituting these back into our differential equation, we obtain:

\frac{1}{3} \int (\frac{1}{y - 3} - \frac{1}{y}) d y = \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2}) d x

The Logarithmic Dance

Integrating both sides results in the following logarithmic expression:

\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln \frac{x - 2}{x + 2} + C

We determine the constant

C

using the initial condition

y (4) = \frac{3}{2}

. Substituting these values:

\frac{1}{3} ln \frac{1.5 - 3}{1.5} = \frac{1}{4} ln \frac{4 - 2}{4 + 2} + C

Since

ln ∣1∣ = 0

, the left side vanishes, leaving

C = - \frac{1}{4} ln (\frac{1}{3}) = \frac{1}{4} ln 3

Final Calculation

Now, we substitute

x = 10

and our value for

C

into the equation:

\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln (\frac{10 - 2}{10 + 2}) + \frac{1}{4} ln 3

Simplifying the right side:

\frac{1}{4} ln (\frac{8}{12}) + \frac{1}{4} ln 3 = \frac{1}{4} ln (\frac{2}{3} \cdot 3) = \frac{1}{4} ln 2

Multiplying by

3

, we find

ln \frac{y - 3}{y} = \frac{3}{4} ln 2 = ln (2^{3/4})

. Given the initial condition, the expression

\frac{y - 3}{y}

must remain negative, so

\frac{y - 3}{y} = - 2^{3/4}

Solving for

y

y - 3 = - y \cdot 2^{3/4} \Rightarrow y (1 + 2^{3/4}) = 3

The final value is:

$y = \frac{3}{1 + 2 ^{3/4}}$

Common Traps & FAQs

Wait, why do we need the absolute value in the natural log? Can't we just ignore it?

That is a great question! The integral of

\frac{1}{u}

ln ∣ u ∣ + C

because the natural logarithm is only defined for positive numbers. When you integrate, you don't know if the expression inside is positive or negative, so the absolute value ensures the math remains valid. However, you cannot just ignore it—you must use your initial condition to determine the sign of the expression inside the absolute value.

Why is the sign negative when we remove the absolute value? How do I know?

This is the classic trap! Look at your initial condition:

y (4) = 1.5

. Plug this into the expression

\frac{y - 3}{y}

. You get

\frac{1.5 - 3}{1.5} = \frac{- 1.5}{1.5} = - 1

. Since the problem states the slope is never zero, the curve cannot cross the lines

y = 0

y = 3

. Because the expression is negative at

x = 4

, it must remain negative for all

x > 2

. Therefore, when you remove the absolute value, you must include a negative sign.

Is there a faster way to solve this without finding the constant

C

Yes, absolutely! You can use definite integration. Instead of finding

C

, you can integrate from the initial state

(x = 4, y = 1.5)

to the final state

(x = 10, y = y_{f ina l})

. The integral on the left would be

\int_{1.5}^{y} \frac{d y}{y ^{2} - 3 y}

and the integral on the right would be

\int_{4}^{10} \frac{d x}{x ^{2} - 4}

. This approach automatically accounts for the constant and is often less prone to algebraic errors.

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of $y (10)$ equals :

Select Answer:

Visualized Solution

The Differential Equation

Variable Separation

Partial Fraction Decomposition (LHS)

Partial Fraction Decomposition (RHS)

Setting Up the Integrals

Integration

Finding the Constant $C$ (Setup)

Calculating $C$

Substituting $x = 10$

Simplifying the RHS

Isolating the Log Term

The Absolute Value Trap

Solving for $y (10)$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Power of Decomposition

The Logarithmic Dance

Final Calculation

Similar Questions

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0$ , $y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (2)$ is ______.

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

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{- 1}{2})$ is equal to :

If the solution $y = y (x)$ of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then $y (0)$ is equal to :

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0$ , $∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

Select Answer:

Visualized Solution

The Differential Equation

Variable Separation

Partial Fraction Decomposition (LHS)

Partial Fraction Decomposition (RHS)

Setting Up the Integrals

Integration

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

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=10

Simplifying the RHS

Isolating the Log Term

The Absolute Value Trap

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(10)

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Power of Decomposition

The Logarithmic Dance

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x3dy+xydx=x2dy+2ydx; y(2)=e and x>1, then y(4) is equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (1+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. Then y(4π​) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

Finding the Constant $C$ (Setup)

Calculating $C$

Substituting $x = 10$

Solving for $y (10)$

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

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{- 1}{2})$ is equal to :

If the solution $y = y (x)$ of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then $y (0)$ is equal to :

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0$ , $∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to: