JEE Advanced 2004

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: A curve 'C' passes through (2, 0) and the slope at (x, y) as $\frac{( x + 1 ) ^{2} + ( y - 3 )}{x + 1}$ . Find the equation of the curve. Find the area bounded by curve and x-axis in fourth quadrant.

Q: Why can't I just separate the variables directly?

Because the variables $x$ and $y$ are not separable. The term $\frac{y-3}{x+1}$ prevents us from isolating $dy$ and $dx$ on opposite sides. This is a linear differential equation, which requires the Integrating Factor method.

Q: I got $C = 3$ instead of $C = -3$. What happened?

Check your signs when integrating $-\frac{1}{x+1}$. The integral is $-\ln(x+1)$, which leads to $I.F. = \frac{1}{x+1}$. If you missed the negative sign, your $I.F.$ would be $(x+1)$, which is incorrect. Always be meticulous with signs in differential equations.

Q: Why is the area $\frac{4}{3}$ and not negative?

Area is a geometric quantity representing magnitude. When the curve lies below the x-axis, the integral is negative. We take the absolute value to represent the physical area, which must always be positive.

Visualized Solution

Identifying the Differential Equation

Given slope at any point $(x, y)$ is $\frac{d y}{d x}$ .
$\frac{d y}{d x} = \frac{( x + 1 ) ^{2} + ( y - 3 )}{x + 1}$
The curve passes through the point $(2, 0)$ .

Rearranging to Linear Form

Split the fraction: $\frac{d y}{d x} = \frac{( x + 1 ) ^{2}}{x + 1} + \frac{y - 3}{x + 1}$
Simplify: $\frac{d y}{d x} = (x + 1) + \frac{y}{x + 1} - \frac{3}{x + 1}$
Rearrange: $\frac{d y}{d x} - (\frac{1}{x + 1}) y = (x + 1) - \frac{3}{x + 1}$

Calculating the Integrating Factor

Compare with $\frac{d y}{d x} + P y = Q$
Here, $P = - \frac{1}{x + 1}$
Integrating Factor (I.F.) $= e^{\int P d x} = e^{\int - \frac{1}{x + 1} d x}$
I.F. $= e^{- l n (x + 1)} = \frac{1}{x + 1}$

Setting up the General Solution

General solution formula: $y \cdot (I.F.) = \int Q \cdot (I.F.) d x$
Substitute I.F. and $Q$ :
$y (\frac{1}{x + 1}) = \int ((x + 1) - \frac{3}{x + 1}) \frac{1}{x + 1} d x$

Simplifying the Integral

Distribute the I.F. inside the integral:
$y (\frac{1}{x + 1}) = \int (\frac{x + 1}{x + 1} - \frac{3}{( x + 1 ) ^{2}}) d x$
$y (\frac{1}{x + 1}) = \int (1 - \frac{3}{( x + 1 ) ^{2}}) d x$

Executing the Integration

Integrate term by term:
$\int 1 d x = x$
$\int - 3 (x + 1)^{- 2} d x = \frac{3}{x + 1}$
Equation becomes: $\frac{y}{x + 1} = x + \frac{3}{x + 1} + C$

Finding the Constant of Integration

The curve passes through $(2, 0)$ .
Substitute $x = 2, y = 0$ :
$\frac{0}{2 + 1} = 2 + \frac{3}{2 + 1} + C$
$0 = 2 + 1 + C ⟹ C = - 3$

Final Equation of the Curve

Substitute $C = - 3$ back into the equation:
$\frac{y}{x + 1} = x + \frac{3}{x + 1} - 3$
Multiply the entire equation by $(x + 1)$ :
$y = x (x + 1) + 3 - 3 (x + 1)$
$y = x^{2} + x + 3 - 3 x - 3 ⟹ y = x^{2} - 2 x$

Analyzing the Bounded Area

We need the area bounded by $y = x^{2} - 2 x$ and the x-axis in the fourth quadrant.
Find x-intercepts by setting $y = 0$ :
$x^{2} - 2 x = 0 ⟹ x (x - 2) = 0$
Intersects at $x = 0$ and $x = 2$ .

Setting up the Area Integral

In the interval $(0, 2)$ , the curve lies below the x-axis ( $y < 0$ ).
Area $A = \int_{0}^{2} (x^{2} - 2 x) d x$
Or, $A = \int_{0}^{2} (0 - y) d x = \int_{0}^{2} (2 x - x^{2}) d x$

Evaluating the Definite Integral

Integrate: $\int (2 x - x^{2}) d x = [x^{2} - \frac{x ^{3}}{3}]_{0}^{2}$
Substitute upper limit $x = 2$ : $(2)^{2} - \frac{( 2 ) ^{3}}{3} = 4 - \frac{8}{3}$
Substitute lower limit $x = 0$ : $0$

Final Area Calculation

$A = 4 - \frac{8}{3}$
$A = \frac{12 - 8}{3} = \frac{4}{3}$ sq. units
Final Answer: Equation is $y = x^{2} - 2 x$ , Area is $\frac{4}{3}$ sq. units.

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

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

By splitting the fraction, we simplify the expression to:

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

Rearranging this into the standard form of a Linear Differential Equation,

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

, we obtain:

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

The Magic of the Integrating Factor

To solve this, we identify

P (x) = - \frac{1}{x + 1}

. The Integrating Factor (I.F.) is defined as

e^{\int P (x) d x}

I . F . = e^{\int - \frac{1}{x + 1} d x} = e^{- l n (x + 1)} = \frac{1}{x + 1}

Multiplying the entire differential equation by this

I . F .

allows us to express the left side as the derivative of a product:

\frac{d}{d x} (y \cdot \frac{1}{x + 1}) = ((x + 1) - \frac{3}{x + 1}) \cdot \frac{1}{x + 1}

Solving the Integral

Simplifying the right side, we get:

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

Integrating both sides with respect to

x

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

Given the curve passes through the point

(2, 0)

, we substitute these values to find

C

0 = 2 + \frac{3}{3} + C \Rightarrow 0 = 3 + C \Rightarrow C = - 3

Thus, the equation of the curve is:

y = (x + 1) (x + \frac{3}{x + 1} - 3) = x^{2} + x + 3 - 3 x - 3 = x^{2} - 2 x

The Final Geometric Triumph

The curve

y = x^{2} - 2 x

intersects the x-axis at

x = 0

and

x = 2

. In the interval

[0, 2]

, the parabola lies below the x-axis.

The area bounded by the curve and the x-axis is given by the absolute value of the definite integral:

A r e a = \int_{0}^{2} (x^{2} - 2 x) d x

Evaluating the integral:

\int_{0}^{2} (x^{2} - 2 x) d x = [\frac{x ^{3}}{3} - x^{2}]_{0}^{2} = (\frac{8}{3} - 4) - 0 = - \frac{4}{3}

Taking the absolute value, the final area is $\frac{4}{3}$ square units.

Common Traps & FAQs

Why can't I just separate the variables directly?

Because the variables

x

and

y

are not separable. The term

\frac{y - 3}{x + 1}

prevents us from isolating

d y

and

d x

on opposite sides. This is a linear differential equation, which requires the Integrating Factor method.

I got

C = 3

instead of

C = - 3

. What happened?

Check your signs when integrating

- \frac{1}{x + 1}

. The integral is

- ln (x + 1)

, which leads to

I . F . = \frac{1}{x + 1}

. If you missed the negative sign, your

I . F .

would be

(x + 1)

, which is incorrect. Always be meticulous with signs in differential equations.

Why is the area

\frac{4}{3}

and not negative?

Area is a geometric quantity representing magnitude. When the curve lies below the x-axis, the integral is negative. We take the absolute value to represent the physical area, which must always be positive.

Animated Solution for Mathematics - Differential Equations: A curve 'C' passes through (2, 0) and the slope at (x, y) as $\frac{( x + 1 ) ^{2} + ( y - 3 )}{x + 1}$ . Find the equation of the curve. Find the area bounded by curve and x-axis in fourth quadrant.

Visualized Solution

Identifying the Differential Equation

Rearranging to Linear Form

Calculating the Integrating Factor

Setting up the General Solution

Simplifying the Integral

Executing the Integration

Finding the Constant of Integration

Final Equation of the Curve

Analyzing the Bounded Area

Setting up the Area Integral

Evaluating the Definite Integral

Final Area Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

Solving the Integral

The Final Geometric Triumph

Similar Questions

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Visualized Solution

Identifying the Differential Equation

Rearranging to Linear Form

Calculating the Integrating Factor

Setting up the General Solution

Simplifying the Integral

Executing the Integration

Finding the Constant of Integration

Final Equation of the Curve

Analyzing the Bounded Area

Setting up the Area Integral

Evaluating the Definite Integral

Final Area Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

Solving the Integral

The Final Geometric Triumph

Similar Questions

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