JEE Main 2021 (February)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: If a curve $y = f (x)$ passes through the point (1,2) and satisfies $x \frac{d y}{d x} + y = b x^{4}$ , then for what value of b, $\int_{1}^{2} f (x) d x = \frac{62}{5}$ ?

Q: How did you know $C$ must be zero in the final step?

This is a classic JEE trap! We arrived at the equation $\frac{31b}{25} + C\ln 2 = \frac{62}{5}$. The right side is a rational number. If $C$ were any non-zero value, $C\ln 2$ would be an irrational number (since $\ln 2$ is transcendental). For the equation to hold, the irrational part must vanish, forcing $C$ to be zero. It is a beautiful logical deduction!

Select Answer:

Visualized Solution

The Problem Setup

We have a curve $y = f (x)$ passing through the point $(1, 2)$ .
It satisfies the differential equation $x \frac{d y}{d x} + y = b x^{4}$ .
We are given the area under the curve from $x = 1$ to $x = 2$ is $\frac{62}{5}$ .

Analyzing the Differential Equation

Given: $x \frac{d y}{d x} + y = b x^{4}$
Notice the left-hand side: $x \frac{d y}{d x} + y \cdot 1$
This is the exact derivative of the product $x y$ .

Rewriting as an Exact Differential

Using the product rule: $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$
Rewrite the equation: $\frac{d}{d x} (x y) = b x^{4}$

Integrating Both Sides

Integrate with respect to $x$ :
$\int \frac{d}{d x} (x y) d x = \int b x^{4} d x$
$x y = \frac{b x ^{5}}{5} + C$

Applying the Initial Condition

The curve passes through the point $(1, 2)$ .
Substitute $x = 1$ and $y = 2$ into $x y = \frac{b x ^{5}}{5} + C$ .
$1 \cdot 2 = \frac{b ( 1 ) ^{5}}{5} + C$

Finding the Constant $C$

Simplify the equation: $2 = \frac{b}{5} + C$
Isolate $C$ : $C = 2 - \frac{b}{5}$

Expressing $f (x)$ Explicitly

We have $x y = \frac{b x ^{5}}{5} + C$
Divide the entire equation by $x$ :
$y = \frac{b x ^{4}}{5} + \frac{C}{x}$
So, $f (x) = \frac{b x ^{4}}{5} + \frac{C}{x}$

Setting up the Definite Integral

We are given the area condition: $\int_{1}^{2} f (x) d x = \frac{62}{5}$
Substitute our expression for $f (x)$ :
$\int_{1}^{2} (\frac{b x ^{4}}{5} + \frac{C}{x}) d x = \frac{62}{5}$

Performing the Integration

Integrate term by term:
$\int \frac{b x ^{4}}{5} d x = \frac{b x ^{5}}{25}$
$\int \frac{C}{x} d x = C ln ∣ x ∣$
Result: $[\frac{b x ^{5}}{25} + C ln ∣ x ∣]_{1}^{2} = \frac{62}{5}$

Applying the Limits

Substitute upper limit ( $x = 2$ ): $\frac{32 b}{25} + C ln 2$
Substitute lower limit ( $x = 1$ ): $\frac{b}{25} + C ln 1$
Subtract: $(\frac{32 b}{25} + C ln 2) - (\frac{b}{25} + 0) = \frac{62}{5}$
Simplify: $\frac{31 b}{25} + C ln 2 = \frac{62}{5}$

Logical Deduction for $C$

We have: $\frac{31 b}{25} + C ln 2 = \frac{62}{5}$
Notice the right side $\frac{62}{5}$ is a rational number.
The term $C ln 2$ is irrational unless $C = 0$ .
Therefore, for the equation to hold with rational $b$ , $C$ must be $0$ .

Solving for $b$

Set $C = 0$ in our earlier relation: $C = 2 - \frac{b}{5}$
$0 = 2 - \frac{b}{5}$
$\frac{b}{5} = 2 ⟹ b = 10$

Final Verification

Let's quickly verify: If $b = 10$ and $C = 0$ , does $\frac{31 b}{25} = \frac{62}{5}$ ?
$\frac{31 ( 10 )}{25} = \frac{310}{25} = \frac{62}{5}$ .
The condition is perfectly satisfied.
Final Answer: $b = 10$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The blue curve

y = f (x)

passes through the point

(1, 2)

. The area under the curve from

x = 1

x = 2

is given as

\frac{62}{5}

The governing differential equation is:

x \frac{d y}{d x} + y = b x^{4}

The Hidden Symmetry

When observing the left-hand side of the equation, we recognize the product rule for differentiation. Specifically, we know that:

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

By applying this identity, the differential equation simplifies to:

\frac{d}{d x} (x y) = b x^{4}

The Integration Journey

Integrating both sides with respect to

x

, we obtain:

x y = \int b x^{4} d x = \frac{b x ^{5}}{5} + C

To determine the constant

C

, we use the anchor point

(1, 2)

. Substituting

x = 1

and

y = 2

into the equation:

1 \cdot 2 = \frac{b ( 1 ) ^{5}}{5} + C \Rightarrow 2 = \frac{b}{5} + C \Rightarrow C = 2 - \frac{b}{5}

The Integral Challenge

The function is defined as

f (x) = \frac{b x ^{4}}{5} + \frac{C}{x}

. We are given that the area under this curve from

1

2

\frac{62}{5}

\int_{1}^{2} (\frac{b x ^{4}}{5} + \frac{C}{x}) d x = \frac{62}{5}

Evaluating the integral term by term:

[\frac{b x ^{5}}{25} + C ln ∣ x ∣]_{1}^{2} = \frac{62}{5}

Substituting the limits of integration:

(\frac{32 b}{25} + C ln 2) - (\frac{b}{25} + C ln 1) = \frac{62}{5}

Since

ln 1 = 0

, the expression simplifies to:

\frac{31 b}{25} + C ln 2 = \frac{62}{5}

The Irrationality Trap

We observe that the right side of the equation,

\frac{62}{5}

, is a rational number. The term

C ln 2

is irrational unless

C = 0

For the equation to hold true, we must have

C = 0

. Substituting this back into our earlier relation

C = 2 - \frac{b}{5}

0 = 2 - \frac{b}{5}

Solving for

b

, we find the final result:

$b = 10$

Common Traps & FAQs

Wait, why can't we just use separation of variables here?

That is a great question! Separation of variables works when you can isolate all

y

terms on one side and all

x

terms on the other. Here, the

y

term is added to

x \frac{d y}{d x}

, not multiplied. This makes it a linear differential equation of the form

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

. While you could use an integrating factor, recognizing the product rule is a much faster, more elegant path!

How did you know

C

must be zero in the final step?

This is a classic JEE trap! We arrived at the equation

\frac{31 b}{25} + C ln 2 = \frac{62}{5}

. The right side is a rational number. If

C

were any non-zero value,

C ln 2

would be an irrational number (since

ln 2

is transcendental). For the equation to hold, the irrational part must vanish, forcing

C

to be zero. It is a beautiful logical deduction!

Why do we need the point (1,2) if we already have the integral condition?

Think of it as two separate constraints. The differential equation gives us a family of curves (the general solution with constant

C

). The point

(1, 2)

anchors the curve to a specific location, allowing us to define

C

in terms of

b

. The integral condition then acts as the final key to unlock the specific value of

b

Animated Solution for Mathematics - Differential Equations: If a curve $y = f (x)$ passes through the point (1,2) and satisfies $x \frac{d y}{d x} + y = b x^{4}$ , then for what value of b, $\int_{1}^{2} f (x) d x = \frac{62}{5}$ ?

Select Answer:

Visualized Solution

The Problem Setup

Analyzing the Differential Equation

Rewriting as an Exact Differential

Integrating Both Sides

Applying the Initial Condition

Finding the Constant $C$

Expressing $f (x)$ Explicitly

Setting up the Definite Integral

Performing the Integration

Applying the Limits

Logical Deduction for $C$

Solving for $b$

Final Verification

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Hidden Symmetry

The Integration Journey

The Integral Challenge

The Irrationality Trap

Similar Questions

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

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

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Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

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Let $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :

Select Answer:

Visualized Solution

The Problem Setup

Analyzing the Differential Equation

Rewriting as an Exact Differential

Integrating Both Sides

Applying the Initial Condition

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

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) Explicitly

Setting up the Definite Integral

Performing the Integration

Applying the Limits

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

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

Final Verification

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Hidden Symmetry

The Integration Journey

The Integral Challenge

The Irrationality Trap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the differential equation, y2dx+(x−y1​)dy=0. If value of y is 1 when x = 1, then the value of x for which y = 2, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dydx​=y1+x−y2​,x(1)=1, then 5x(2) 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 be the solution of the differential equation (1−x2)dy=[xy+(x3+2)1−x2​]dx,−1<x<1 and y(0)=0. If ∫−1/21/2​1−x2​y(x)dx=k, then k−1 is equal to :

Finding the Constant $C$

Expressing $f (x)$ Explicitly

Logical Deduction for $C$

Solving for $b$

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Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :