JEE Main 2021 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x - x^{3}) d y = (y + y x^{2} - 3 x^{4}) d x, x > 2$ . If $y (3) = 3$ , then $y (4)$ is equal to:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Given: $(x - x^{3}) d y = (y + y x^{2} - 3 x^{4}) d x$
Condition: $y (3) = 3$
Goal: Find $y (4)$

Expand and Distribute Terms

Expanding the terms:
$x d y - x^{3} d y = y d x + y x^{2} d x - 3 x^{4} d x$

Group Related Differentials

Rearranging to group $x d y$ and $y d x$ :
$x d y - y d x = x^{3} d y + y x^{2} d x - 3 x^{4} d x$

Divide by $x^{2}$ for Quotient Form

Dividing the entire equation by $x^{2}$ :
$\frac{x d y - y d x}{x ^{2}} = \frac{x ^{3} d y}{x ^{2}} + \frac{y x ^{2} d x}{x ^{2}} - \frac{3 x ^{4} d x}{x ^{2}}$
$\frac{x d y - y d x}{x ^{2}} = x d y + y d x - 3 x^{2} d x$

Identify Exact Differentials

Recognizing the exact differentials:
Left Side: $\frac{x d y - y d x}{x ^{2}} = d (\frac{y}{x})$
Right Side: $x d y + y d x = d (x y)$
Result: $d (\frac{y}{x}) = d (x y) - d (x^{3})$

Integrate Both Sides

Integrating both sides:
$\int d (\frac{y}{x}) = \int d (x y) - \int d (x^{3})$
$\frac{y}{x} = x y - x^{3} + C$

Apply Initial Condition

Substitute $x = 3$ and $y = 3$ :
$\frac{3}{3} = (3) (3) - 3^{3} + C$
$1 = 9 - 27 + C$

Solve for Constant C

$1 = - 18 + C ⟹ C = 19$
The particular solution is:
$\frac{y}{x} = x y - x^{3} + 19$
$y = \frac{x ( x ^{3} - 19 )}{x ^{2} - 1}$

Find y when x = 4

Substitute $x = 4$ :
$\frac{y}{4} = 4 y - 4^{3} + 19$
$\frac{y}{4} = 4 y - 64 + 19$

Simplify the Equation

Simplify constants:
$\frac{y}{4} = 4 y - 45$
Multiply by 4:
$y = 16 y - 180$

Final Calculation for y(4)

Rearranging:
$180 = 16 y - y$
$15 y = 180$
$y = \frac{180}{15} = 12$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We are tasked with solving the differential equation:

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

subject to the initial condition

y (3) = 3

. Our goal is to determine the value of

y (4)

First, we expand the terms to observe the structure of the equation:

x d y - x^{3} d y = y d x + y x^{2} d x - 3 x^{4} d x

Rearranging the Terms

To simplify, we group the terms involving

x

and

y

strategically. We bring

x d y

and

y d x

together on one side:

x d y - y d x = x^{3} d y + y x^{2} d x - 3 x^{4} d x

We recognize that the left side resembles the numerator of the derivative of a quotient, specifically

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

. To facilitate this, we divide the entire equation by

x^{2}

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

The Master Equation

The left side is now the exact differential

d (\frac{y}{x})

. The right side can be rewritten as:

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

Integrating both sides of the equation yields:

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

Final Calculation

We apply the initial condition

y (3) = 3

to solve for the constant

C

\frac{3}{3} = (3) (3) - (3)^{3} + C

1 = 9 - 27 + C

1 = - 18 + C \Rightarrow C = 19

The general solution is therefore:

\frac{y}{x} = x y - x^{3} + 19

To find

y (4)

, we substitute

x = 4

into the equation:

\frac{y}{4} = 4 y - (4)^{3} + 19

\frac{y}{4} = 4 y - 64 + 19

\frac{y}{4} = 4 y - 45

Multiplying by 4, we get

y = 16 y - 180

, which simplifies to

15 y = 180

. Thus, the final result is: $y (4) = 12$

Common Traps & FAQs

Why did we divide by

x^{2}

? How was I supposed to guess that?

That is a fantastic question! In differential equations, we often look for 'exact differentials.' When you see terms like

x d y

and

y d x

appearing together, it is a massive hint that you are dealing with the product rule

d (x y) = x d y + y d x

or the quotient rule

d (y / x) = (x d y - y d x) / x^{2}

. By dividing by

x^{2}

, we forced the equation into a form where the left side became the derivative of a simple quotient. It is not guessing; it is pattern recognition!

Can't I just use the standard linear differential equation form

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

You absolutely could! If you rearrange the equation into the form

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

, you would find an integrating factor. However, that path involves complex integration that is prone to errors. The 'exact differential' method we used is the elegant, 'JEE-style' shortcut. It turns a potentially messy integration problem into a simple algebraic one. Always look for the elegant path first!

What if I didn't see the exact differential? Is there a backup plan?

If you don't spot the exact differential, don't panic! You can always fall back on the standard linear form

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

. It will lead to the same result, just with more steps and more room for calculation errors. The key is to keep practicing these problems so that the structure of exact differentials becomes second nature to you. You've got this!

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x - x^{3}) d y = (y + y x^{2} - 3 x^{4}) d x, x > 2$ . If $y (3) = 3$ , then $y (4)$ is equal to:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Expand and Distribute Terms

Group Related Differentials

Divide by $x^{2}$ for Quotient Form

Identify Exact Differentials

Integrate Both Sides

Apply Initial Condition

Solve for Constant C

Find y when x = 4

Simplify the Equation

Final Calculation for y(4)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Rearranging the Terms

The Master Equation

Final Calculation

Similar Questions

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1$ , with $y (2) = - 2$ . Then $y (3)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x, y (0) = 1$ . Then $\int_{- 3}^{3} y (x) d x$ is :

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

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

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Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Select Answer:

Visualized Solution

Analyze the Differential Equation

Expand and Distribute Terms

Group Related Differentials

Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2 for Quotient Form

Identify Exact Differentials

Integrate Both Sides

Apply Initial Condition

Solve for Constant C

Find y when x = 4

Simplify the Equation

Final Calculation for y(4)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Rearranging the Terms

The Master Equation

Final Calculation

Similar Questions

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Divide by $x^{2}$ for Quotient Form

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1$ , with $y (2) = - 2$ . Then $y (3)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x, y (0) = 1$ . Then $\int_{- 3}^{3} y (x) d x$ is :

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

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

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 $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

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 $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

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