JEE Main 2018 (15 April Evening)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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 :-

Q: Wait, why can't we just separate the variables $x$ and $y$ directly from the start?

That is a fantastic question! If you try to isolate $dx$ and $dy$ terms, you will find that the $x$ and $y$ variables are 'entangled' in a way that makes direct separation impossible. The term $2xydy$ prevents you from grouping all $y$ terms with $dy$ and all $x$ terms with $dx$. That is exactly why we use the homogeneous substitution $y=vx$—it acts as a mathematical 'un-entangler' that allows us to separate the variables.

Q: How do I know for sure that the final curve is a circle and not a hyperbola?

Great intuition! The general solution we derived is $x^2 + y^2 = Cx$. When you rearrange this into the form $x^2 - Cx + y^2 = 0$ and complete the square for $x$, you get $(x - \frac{C}{2})^2 + y^2 = (\frac{C}{2})^2$. This is the standard equation of a circle with center $(\frac{C}{2}, 0)$ and radius $\frac{C}{2}$. Since our constant $C$ turned out to be $2$, the equation becomes $(x-1)^2 + y^2 = 1$, which is definitively a circle of radius one.

Select Answer:

Progressive Hints

Hint 1

The Spark: Look closely at the differential equation

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

. Notice that every term has a combined degree of two. This is the hallmark of a homogeneous differential equation. Visualize the symmetry; this is your first clue that the solution will involve a circular or hyperbolic structure.

Hint 2

The Arsenal: When you encounter a homogeneous equation, your most powerful tool is the substitution

y = v x

. This transformation is designed to collapse the complexity of the equation into a separable form. Remember, if you substitute

y

, you must also substitute

d y

using the product rule:

d y = v d x + x d v

Hint 3

The Execution: After substituting

y = v x

and simplifying, you will arrive at the separable form

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

. Now, separate the variables

v

and

x

to get

\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{1}{x} d x

. Integrate both sides to find the general solution.

Visualized Solution

Analyze the DE: $(x^{2} - y^{2}) d x + 2 x y d y = 0$

Given DE: $(x^{2} - y^{2}) d x + 2 x y d y = 0$
Initial Condition: Passes through $(1, 1)$
Identify the type: Every term has a combined degree of 2, suggesting it is a Homogeneous Differential Equation.

Rearrange for $\frac{d y}{d x}$

Rearranging the terms: $2 x y d y = - (x^{2} - y^{2}) d x$
Isolating the derivative: $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$

Substitution $y = v x$

Let $y = v x$
Differentiating with respect to $x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Substitute into the DE

Substituting into the DE: $v + x \frac{d v}{d x} = \frac{( v x ) ^{2} - x ^{2}}{2 x ( v x )}$
Simplifying the right side: $v + x \frac{d v}{d x} = \frac{x ^{2} ( v ^{2} - 1 )}{2 x ^{2} v}$
Canceling $x^{2}$ : $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}$

Simplify the Expression

Isolating $x \frac{d v}{d x}$ : $x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v$
Taking common denominator: $x \frac{d v}{d x} = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v}$
Final simplified form: $x \frac{d v}{d x} = \frac{- ( 1 + v ^{2} )}{2 v}$

Separate the Variables

Separating variables $v$ and $x$ :
$\frac{2 v}{1 + v ^{2}} d v = - \frac{1}{x} d x$

Integrate Both Sides

Integrating both sides: $\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{1}{x} d x$
$ln (1 + v^{2}) = - ln (x) + ln (C)$

Combine Logarithmic Terms

Combining logarithmic terms: $ln (1 + v^{2}) + ln (x) = ln (C)$
$ln (x (1 + v^{2})) = ln (C)$
Removing logs: $x (1 + v^{2}) = C$

Back-substitution of $v$

Substitute $v = \frac{y}{x}$ : $x (1 + \frac{y ^{2}}{x ^{2}}) = C$
Simplifying: $x (\frac{x ^{2} + y ^{2}}{x ^{2}}) = C$
General Equation: $x^{2} + y^{2} = C x$

Apply Initial Condition $(1, 1)$

The curve passes through $(1, 1)$ .
Substitute $x = 1, y = 1$ into $x^{2} + y^{2} = C x$ :
$1^{2} + 1^{2} = C (1) ⟹ C = 2$

Complete the Square

Substituting $C = 2$ : $x^{2} + y^{2} = 2 x$
Rearranging: $x^{2} - 2 x + y^{2} = 0$
Completing the square for $x$ : $(x^{2} - 2 x + 1) + y^{2} = 1$
$(x - 1)^{2} + y^{2} = 1$

Identify the Curve

Standard form of a circle: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Comparing: Center $(h, k) = (1, 0)$ and $r^{2} = 1 ⟹ r = 1$
The curve is a circle of radius one.

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The differential equation provided is

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

. At first glance, it appears complex, but it possesses a beautiful, symmetric structure.

The key to unlocking this problem lies in recognizing its homogeneity. Notice that in the term

x^{2}

, the degree is two; in

y^{2}

, the degree is two; and in

2 x y

, the combined degree of

x

and

y

is also two.

This uniform degree is a massive signal. It indicates that the equation scales perfectly, allowing us to use the substitution

y = v x

to simplify the expression.

The Transformation

We begin by rearranging the equation to isolate the derivative

\frac{d y}{d x}

. By moving the

d x

term to the right and dividing, we obtain:

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

Now, we introduce our substitution

y = v x

. Differentiating this with respect to

x

using the product rule gives us:

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

Substituting these into our equation transforms the right side into:

\frac{( v x ) ^{2} - x ^{2}}{2 x ( v x )}

Notice the mathematical elegance here: the

x^{2}

terms in the numerator and denominator cancel out completely. This leaves us with the simplified expression:

\frac{v ^{2} - 1}{2 v}

The Path to Integration

We are now left with the equation

v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}

. Subtracting

v

from both sides and finding a common denominator leads us to:

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

The variables are now ready to be separated. We bring all

v

terms to the left and all

x

terms to the right, resulting in:

\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{1}{x} d x

This integral is a classic form. Since the numerator is the derivative of the denominator, the left side integrates to

ln (1 + v^{2})

. The right side is simply

- ln (x) + ln (C)

The Geometric Reveal

Combining the logarithms gives us

ln (x (1 + v^{2})) = ln (C)

, which simplifies to

x (1 + v^{2}) = C

. Substituting back

v = \frac{y}{x}

, we get:

x (1 + \frac{y ^{2}}{x ^{2}}) = C \Rightarrow x^{2} + y^{2} = C x

We are given that the curve passes through the point

(1, 1)

. Substituting these values gives

1^{2} + 1^{2} = C (1)

, which means

C = 2

Our final equation is

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

. Rearranging this into the standard form:

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

We see the unmistakable signature of a circle with center $(1, 0)$ and radius $1$ . You have successfully navigated the differential equation to reveal a perfect geometric truth.

Common Traps & FAQs

Wait, why can't we just separate the variables

x

and

y

directly from the start?

That is a fantastic question! If you try to isolate

d x

and

d y

terms, you will find that the

x

and

y

variables are 'entangled' in a way that makes direct separation impossible. The term

2 x y d y

prevents you from grouping all

y

terms with

d y

and all

x

terms with

d x

. That is exactly why we use the homogeneous substitution

y = v x

—it acts as a mathematical 'un-entangler' that allows us to separate the variables.

How do I know for sure that the final curve is a circle and not a hyperbola?

Great intuition! The general solution we derived is

x^{2} + y^{2} = C x

. When you rearrange this into the form

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

and complete the square for

x

, you get

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

. This is the standard equation of a circle with center

(\frac{C}{2}, 0)

and radius

\frac{C}{2}

. Since our constant

C

turned out to be

2

, the equation becomes

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

, which is definitively a circle of radius one.

Why did we choose the natural log of

C

as our constant of integration?

That is a clever algebraic shortcut! When you integrate both sides, you get

ln (1 + v^{2}) = - ln (x) + K

. If you leave the constant as just

K

, you would have to deal with

e^{K}

later, which can be messy. By writing the constant as

ln (C)

, you can use the logarithmic property

ln (A) + ln (B) = ln (A B)

to combine everything into a single logarithm, allowing you to 'drop' the logs from both sides much more cleanly. It is all about making the algebra work for you, not against you!

Animated Solution for Mathematics - Differential Equations: 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 :-

Select Answer:

Visualized Solution

Analyze the DE: $(x^{2} - y^{2}) d x + 2 x y d y = 0$

Rearrange for $\frac{d y}{d x}$

Substitution $y = v x$

Substitute into the DE

Simplify the Expression

Separate the Variables

Integrate Both Sides

Combine Logarithmic Terms

Back-substitution of $v$

Apply Initial Condition $(1, 1)$

Complete the Square

Identify the Curve

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Path to Integration

The Geometric Reveal

Similar Questions

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 :

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Animated Solution for Mathematics - Differential Equations: .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 :-

Select Answer:

Visualized Solution

Analyze the DE: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x2−y2)dx+2xydy=0

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

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=vx

Substitute into the DE

Simplify the Expression

Separate the Variables

Integrate Both Sides

Combine Logarithmic Terms

Back-substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1)

Complete the Square

Identify the Curve

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Path to Integration

The Geometric Reveal

Similar Questions

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A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

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Animated Solution for Mathematics - Differential Equations: 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 :-

Analyze the DE: $(x^{2} - y^{2}) d x + 2 x y d y = 0$

Rearrange for $\frac{d y}{d x}$

Substitution $y = v x$

Back-substitution of $v$

Apply Initial Condition $(1, 1)$

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 :

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 solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

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

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

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 the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :