JEE Main 2021 (16 March Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $C_{1}$ be the curve obtained by the solution of differential equation $2 x y \frac{d y}{d x} = y^{2} - x^{2}, x > 0$ & Let the curve $C_{2}$ be the solution of $\frac{2 x y}{x ^{2} - y ^{2}} = \frac{d y}{d x}$ . If both the curves pass through $(1, 1)$ , then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to :

Q: How do I know for sure that the area is symmetric?

Look at the equations: $C_1: (x-1)^2 + y^2 = 1$ and $C_2: x^2 + (y-1)^2 = 1$. If you swap $x$ and $y$ in the equation for $C_1$, you get the equation for $C_2$. This means the curves are reflections of each other across the line $y = x$. Since the line of symmetry passes through the intersection points $(0, 0)$ and $(1, 1)$, the area must be symmetric.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Observe that both differential equations are homogeneous of degree 2. This is a classic indicator that the substitution

y = v x

will simplify the expressions significantly. Also, visualize the resulting curves; they are not just arbitrary functions, but geometric shapes.

Hint 2

Hint 2 (The Arsenal): Once you perform the substitution

y = v x

and separate the variables, you will find that both equations lead to the form of a circle. Remember to use the condition that both curves pass through

(1, 1)

to determine the constants of integration.

Hint 3

Hint 3 (The Execution): After solving, you will find

C_{1}

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

and

C_{2}

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

. The area enclosed is the intersection of two unit circles. Use the symmetry about the line

y = x

to calculate the area of one segment and double it.

Visualized Solution

Analyzing Curve $C_{1}$

Given DE for $C_{1}$ : $2 x y \frac{d y}{d x} = y^{2} - x^{2}$
Rearranging: $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$
This is a Homogeneous Differential Equation of degree 2.

Substitution Method for $C_{1}$

Let $y = v x$ , which implies $\frac{d y}{d x} = v + x \frac{d v}{d x}$
Substituting into the DE: $v + x \frac{d v}{d x} = \frac{v ^{2} x ^{2} - x ^{2}}{2 x ( v x )}$
Factoring out $x^{2}$ : $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}$

Separating Variables for $C_{1}$

Isolating the derivative term: $x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v$
Simplifying the RHS: $x \frac{d v}{d x} = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v} = - \frac{v ^{2} + 1}{2 v}$
Separating variables: $\frac{2 v}{v ^{2} + 1} d v = - \frac{1}{x} d x$

Integrating to find $C_{1}$

Integrating both sides: $\int \frac{2 v}{v ^{2} + 1} d v = - \int \frac{1}{x} d x$
Result: $ln (v^{2} + 1) = - ln x + ln c$
Combining logarithms: $ln (x (v^{2} + 1)) = ln c \Rightarrow x (v^{2} + 1) = c$

Final Equation of $C_{1}$

Substitute back $v = \frac{y}{x}$ : $x (\frac{y ^{2}}{x ^{2}} + 1) = c \Rightarrow x^{2} + y^{2} = c x$
The curve passes through $(1, 1)$ , so $1^{2} + 1^{2} = c (1) \Rightarrow c = 2$
Equation of $C_{1}$ : $x^{2} + y^{2} - 2 x = 0$ , which is a circle $(x - 1)^{2} + y^{2} = 1$

Analyzing Curve $C_{2}$

Given DE for $C_{2}$ : $\frac{d y}{d x} = \frac{2 x y}{x ^{2} - y ^{2}}$
This is also a Homogeneous Differential Equation.
We apply the same substitution: $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substitution Method for $C_{2}$

Substituting into the DE: $v + x \frac{d v}{d x} = \frac{2 x ( v x )}{x ^{2} - v ^{2} x ^{2}}$
Factoring out $x^{2}$ : $v + x \frac{d v}{d x} = \frac{2 v}{1 - v ^{2}}$
Isolating the derivative: $x \frac{d v}{d x} = \frac{2 v}{1 - v ^{2}} - v = \frac{v + v ^{3}}{1 - v ^{2}}$

Integrating to find $C_{2}$

Separating variables: $\frac{1 - v ^{2}}{v ( 1 + v ^{2} )} d v = \frac{1}{x} d x$
Using partial fractions: $(\frac{1}{v} - \frac{2 v}{1 + v ^{2}}) d v = \frac{1}{x} d x$
Integrating: $ln v - ln (1 + v^{2}) = ln x + ln c \Rightarrow \frac{v}{1 + v ^{2}} = c x$

Final Equation of $C_{2}$

Substitute back $v = \frac{y}{x}$ : $\frac{y / x}{1 + y ^{2} / x ^{2}} = c x \Rightarrow \frac{x y}{x ^{2} + y ^{2}} = c x \Rightarrow x^{2} + y^{2} = \frac{y}{c}$
Let $\frac{1}{c} = k$ , so $x^{2} + y^{2} = k y$
Passes through $(1, 1)$ : $1^{2} + 1^{2} = k (1) \Rightarrow k = 2$
Equation of $C_{2}$ : $x^{2} + y^{2} - 2 y = 0$ , which is a circle $x^{2} + (y - 1)^{2} = 1$

Visualizing the Enclosed Area

$C_{1}$ is a circle centered at $(1, 0)$ with radius $1$ .
$C_{2}$ is a circle centered at $(0, 1)$ with radius $1$ .
They intersect at $(0, 0)$ and $(1, 1)$ .
The enclosed area is symmetric about the line $y = x$ .

Setting up the Area Calculation

The area is divided into two equal parts by the line $y = x$ .
Area = $2 \times$ (Area between $C_{1}$ and $y = x$ )
Alternatively, use geometry: Area = $2 \times (Area of Quarter Circle - Area of Triangle)$

Calculating the Area

Area of the quarter circle of radius 1 is $\frac{π ( 1 ) ^{2}}{4} = \frac{π}{4}$
Area of the right triangle formed by $(0, 0), (1, 0), (1, 1)$ is $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$
Area of one segment = $\frac{π}{4} - \frac{1}{2}$

Final Result

Total Enclosed Area = $2 \times (\frac{π}{4} - \frac{1}{2})$
Total Area = $\frac{π}{2} - 1$
The correct option is (2).

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two mysterious curves,

C_{1}

and

C_{2}

. They seem complex, defined by differential equations that might make your heart race.

Today, we are going to peel back the layers of these equations and reveal the elegant geometry hiding underneath.

Decoding the First Curve

We start with the differential equation for

C_{1}

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

This is a homogeneous differential equation of degree two. When you see terms like

x^{2}

y^{2}

, and

x y

all mixed together, that is your cue to use the substitution

y = v x

By substituting

y = v x

and

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

, the equation transforms into something much more manageable. After some algebraic simplification, we separate the variables and integrate.

The result is a beautiful, simple circle:

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

It is centered at

(1, 0)

with a radius of

1

Unveiling the Second Curve

Now, we turn our attention to

C_{2}

, defined by:

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

Again, we apply the same substitution

y = v x

. The algebra here is just as satisfying.

As we work through the steps, the variables separate, and we find ourselves with another circle:

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

This circle is centered at

(0, 1)

with a radius of

1

The Geometric Revelation

Now, look at the two circles.

C_{1}

is centered at

(1, 0)

and

C_{2}

is centered at

(0, 1)

. They both pass through the origin

(0, 0)

and the point

(1, 1)

The region enclosed by these two circles is a lens-shaped area. Because the equations are symmetric with respect to

x

and

y

, the line

y = x

perfectly bisects this area.

Instead of performing a terrifying integral, we can use simple geometry. The area of the region is twice the area of the circular segment formed by the chord connecting

(0, 0)

and

(1, 1)

in circle

C_{1}

Final Calculation

The area of this segment is simply the area of the quarter circle minus the area of the right-angled triangle.

The quarter circle has an area of:

\frac{π ( 1 ) ^{2}}{4} = \frac{π}{4}

The triangle has an area of:

\frac{1}{2} \times 1 \times 1 = \frac{1}{2}

Thus, the area of one segment is

\frac{π}{4} - \frac{1}{2}

. Multiplying this by two gives us our final, elegant answer:

\frac{π}{2} - 1

Isn't it wonderful how complex calculus can collapse into such a simple, beautiful geometric truth?

Common Traps & FAQs

Wait, why do we reject the negative root here? Or do we?

In this problem, we are dealing with the geometry of circles. When we solve the differential equations, we get the general form

x^{2} + y^{2} - 2 x = 0

and

x^{2} + y^{2} - 2 y = 0

. These are standard circle equations. We don't reject roots; rather, we identify the full path of the curve. The condition that they pass through

(1, 1)

uniquely defines these specific circles.

Why can't we just use simple integration of the difference of the functions?

That is a great question! You could, but these are implicit functions (circles), not simple functions of

x

. To integrate, you would have to express

y

in terms of

x

y = 1 \pm 1 - (x - 1)^{2}

, which introduces square roots and makes the integration much more complex. Using the geometric property of circular segments is much more elegant and less prone to calculation errors.

How do I know for sure that the area is symmetric?

Look at the equations:

C_{1} : (x - 1)^{2} + y^{2} = 1

and

C_{2} : x^{2} + (y - 1)^{2} = 1

. If you swap

x

and

y

in the equation for

C_{1}

, you get the equation for

C_{2}

. This means the curves are reflections of each other across the line

y = x

. Since the line of symmetry passes through the intersection points

(0, 0)

and

(1, 1)

, the area must be symmetric.

Select Answer:

Visualized Solution

Analyzing Curve $C_{1}$

Substitution Method for $C_{1}$

Separating Variables for $C_{1}$

Integrating to find $C_{1}$

Final Equation of $C_{1}$

Analyzing Curve $C_{2}$

Substitution Method for $C_{2}$

Integrating to find $C_{2}$

Final Equation of $C_{2}$

Visualizing the Enclosed Area

Setting up the Area Calculation

Calculating the Area

Final Result

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Decoding the First Curve

Unveiling the Second Curve

The Geometric Revelation

Final Calculation

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $x d y - y d x = x^{2} - y^{2} d x, x \geq 1$ , with $y (1) = 0$ . If the area bounded by the line $x = 1, x = e^{π}, y = 0$ and $y = y (x)$ is $α e^{2 π} + β$ , then the value of $10 (α + β)$ is equal to ____.

Let $y = y (x)$ be the solution of the differential equation $x tan (\frac{y}{x}) d y = (y tan (\frac{y}{x}) - x) d x$ , $- 1 \leq x \leq 1, y (\frac{1}{2}) = \frac{π}{6}$ . Then the area of the region bounded by the curves $x = 0, x = \frac{1}{2}$ and $y = y (x)$ in the upper half plane is:

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

The curve amongst the family of curves, represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ is :

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point (1, 1) is :-

If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point $(2, 1)$ is $tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣$ , then $5 β + α$ is equal to

Let the solution curve $y = y (x)$ of the differential equation, $[\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] y$ pass through the points $(1, 0)$ and $(2 α, α), α > 0$ . Then $α$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . then $∣ (y (2))^{3} - 12 y (2) ∣$ is equal to:

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

The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to

Select Answer:

Visualized Solution

Analyzing Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Substitution Method for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Separating Variables for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Integrating to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Final Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Analyzing Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Substitution Method for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Integrating to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Final Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Visualizing the Enclosed Area

Setting up the Area Calculation

Calculating the Area

Final Result

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Decoding the First Curve

Unveiling the Second Curve

The Geometric Revelation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a curve y=f(x), passing through the point (1,2), is the solution of the differential equation, 2x2dy=(2xy+y2)dx, then f(21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve amongst the family of curves, represented by the differential equation, (x2−y2)dx+2xydy=0 which passes through (1,1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve satisfying the differential equation, (x2−y2)dx+2xydy=0 and passing through the point (1, 1) is :-

.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 (3y2−5x2)ydx+2x(x2−y2)dy=0 such that y(1)=1. then ∣(y(2))3−12y(2)∣ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve of the differential equation xdy−ydx=x2+y2​dx,x>0,y(1)=0, be y=y(x). Then y(3) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of primitive integral equation (x2+y2)dy=xydx is y=y(x). If y(1)=1 and y(x0​)=e, then x0​ is equal to

Analyzing Curve $C_{1}$

Substitution Method for $C_{1}$

Separating Variables for $C_{1}$

Integrating to find $C_{1}$

Final Equation of $C_{1}$

Analyzing Curve $C_{2}$

Substitution Method for $C_{2}$

Integrating to find $C_{2}$

Final Equation of $C_{2}$

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

The curve amongst the family of curves, represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ is :

The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point (1, 1) is :-

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . then $∣ (y (2))^{3} - 12 y (2) ∣$ is equal to:

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

The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to