JEE Main 2023 (24 January Shift 2)

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Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

Q: How can I be absolutely sure that the equation is homogeneous?

To check for homogeneity, replace $x$ with $kx$ and $y$ with $ky$ in every term of the equation. If you can factor out $k^n$ from every term (where $n$ is the degree), the equation is homogeneous of degree $n$. In this problem, every term has a degree of 2, so it is perfectly homogeneous.

Select Answer:

Visualized Solution

The Given Equation

Given differential equation: $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0$
Initial condition: $y (1) = 1$
Goal: Find $6 y^{2} (e)$

Rearranging the Terms

Rearrange to isolate $\frac{d y}{d x}$ :
$(x^{2} - 3 y^{2}) d x = - 3 x y d y$
$\frac{d y}{d x} = \frac{- ( x ^{2} - 3 y ^{2} )}{3 x y}$
$\frac{d y}{d x} = \frac{3 y ^{2} - x ^{2}}{3 x y}$

Identifying Homogeneity

Check the degree of each term:
Numerator: $3 y^{2}$ (deg 2), $- x^{2}$ (deg 2)
Denominator: $3 x y$ (deg 2)
Since all terms have degree 2, it is a Homogeneous Differential Equation.
Simplify: $\frac{d y}{d x} = \frac{y}{x} - \frac{x}{3 y}$

The Substitution $y = v x$

Let $y = v x$
This implies $v = \frac{y}{x}$

Differentiating the Substitution

Differentiate $y = v x$ with respect to $x$ using the product rule:
$\frac{d y}{d x} = v \cdot \frac{d}{d x} (x) + x \cdot \frac{d v}{d x}$
$\frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting into the Equation

Substitute $\frac{d y}{d x} = v + x \frac{d v}{d x}$ and $y = v x$ into the simplified DE:
$v + x \frac{d v}{d x} = v - \frac{x}{3 ( v x )}$
$v + x \frac{d v}{d x} = v - \frac{1}{3 v}$

Simplifying the Expression

Cancel $v$ from both sides:
$x \frac{d v}{d x} = - \frac{1}{3 v}$

Separating $v$ and $x$

Separate the variables $v$ and $x$ :
$3 v d v = - \frac{d x}{x}$

Integrating Both Sides

Integrate both sides:
$\int 3 v d v = - \int \frac{1}{x} d x$
$\frac{3 v ^{2}}{2} = - ln x + C$

Back-Substitution

Substitute $v = \frac{y}{x}$ back into the equation:
$\frac{3 ( y / x ) ^{2}}{2} = - ln x + C$
$\frac{3 y ^{2}}{2 x ^{2}} = - ln x + C$

Applying the Initial Condition

Use the condition $y (1) = 1$ :
$\frac{3 ( 1 ) ^{2}}{2 ( 1 ) ^{2}} = - ln (1) + C$
$\frac{3}{2} = 0 + C ⟹ C = \frac{3}{2}$

The Particular Solution

Substitute $C = \frac{3}{2}$ back into the general solution:
$\frac{3 y ^{2}}{2 x ^{2}} = - ln x + \frac{3}{2}$

Evaluating at $x = e$

Substitute $x = e$ into the equation:
$\frac{3 y ^{2} ( e )}{2 e ^{2}} = - ln (e) + \frac{3}{2}$
$\frac{3 y ^{2} ( e )}{2 e ^{2}} = - 1 + \frac{3}{2}$
$\frac{3 y ^{2} ( e )}{2 e ^{2}} = \frac{1}{2}$

Final Calculation

Solve for $3 y^{2} (e)$ :
$3 y^{2} (e) = e^{2}$
Multiply by 2 to find $6 y^{2} (e)$ :
$6 y^{2} (e) = 2 e^{2}$

Summary and Conclusion

Key Takeaway: Homogeneous DEs of the form $\frac{d y}{d x} = f (\frac{y}{x})$ are solved using $y = v x$ .
Final Result: $6 y^{2} (e) = 2 e^{2}$
Correct Option: [2]

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

(x^{2} - 3 y^{2}) d x + 3 x y d y = 0

. To begin, we rearrange the terms to isolate the derivative

\frac{d y}{d x}

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

Observe that every term in the numerator and denominator is of degree 2. This confirms that the equation is a Homogeneous Differential Equation, which allows us to simplify the problem using a specific substitution.

The Magic of Substitution

We employ the substitution

y = v x

, where

v

is a function of

x

. Differentiating this with respect to

x

using the product rule yields:

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

Substituting these into our differential equation, the

v

terms simplify, leading to a separable equation:

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

x \frac{d v}{d x} = \frac{3 v ^{2} - 1}{3 v} - v = \frac{3 v ^{2} - 1 - 3 v ^{2}}{3 v} = - \frac{1}{3 v}

The Final Integration

We now separate the variables

v

and

x

to solve the equation:

3 v d v = - \frac{d x}{x}

Integrating both sides, we obtain:

\int 3 v d v = - \int \frac{1}{x} d x \Rightarrow \frac{3 v ^{2}}{2} = - ln ∣ x ∣ + C

Substituting

v = \frac{y}{x}

back into the expression, we get the general solution:

\frac{3 y ^{2}}{2 x ^{2}} = - ln ∣ x ∣ + C

Applying Conditions and Final Calculation

Using the initial condition

y (1) = 1

, we substitute these values to find the constant

C

\frac{3 ( 1 ) ^{2}}{2 ( 1 ) ^{2}} = - ln (1) + C \Rightarrow C = \frac{3}{2}

The particular solution is therefore

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

. To find

6 y^{2} (e)

, we substitute

x = e

\frac{3 y ^{2} ( e )}{2 e ^{2}} = - ln (e) + \frac{3}{2} = - 1 + 1.5 = 0.5

Solving for

3 y^{2} (e)

, we get

3 y^{2} (e) = e^{2}

. Multiplying by 2, we arrive at the final result:

$6 y^{2} (e) = 2 e^{2}$

Common Traps & FAQs

Why do we use the substitution

y = v x

instead of something else?

Great question! In homogeneous differential equations, the ratio

\frac{y}{x}

is the key to unlocking the problem. By setting

y = v x

, we transform the equation into one where the variables

v

and

x

can be separated easily. It is a standard, powerful technique that turns a complex, coupled equation into a much simpler, solvable one.

How can I be absolutely sure that the equation is homogeneous?

To check for homogeneity, replace

x

with

k x

and

y

with

k y

in every term of the equation. If you can factor out

k^{n}

from every term (where

n

is the degree), the equation is homogeneous of degree

n

. In this problem, every term has a degree of 2, so it is perfectly homogeneous.

What if I get stuck during the integration step?

Don't panic! Integration is just the reverse of differentiation. Once you have separated the variables, you will have an expression like

\int 3 v d v = - \int \frac{1}{x} d x

. The integral of

v

\frac{v ^{2}}{2}

, and the integral of

\frac{1}{x}

ln ∣ x ∣

. Just remember to add the constant of integration

C

and use the initial condition

y (1) = 1

to find its value.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

Select Answer:

Visualized Solution

The Given Equation

Rearranging the Terms

Identifying Homogeneity

The Substitution $y = v x$

Differentiating the Substitution

Substituting into the Equation

Simplifying the Expression

Separating $v$ and $x$

Integrating Both Sides

Back-Substitution

Applying the Initial Condition

The Particular Solution

Evaluating at $x = e$

Final Calculation

Summary and Conclusion

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Magic of Substitution

The Final Integration

Applying Conditions and Final Calculation

Similar Questions

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ 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:

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

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If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

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

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

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}, y (1) = 1$ . If for some $n \in N, y (2) \in [n - 1, n)$ , then $n$ is equal to

Animated Solution for Mathematics - Differential Equations: .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 (x2−3y2)dx+3xydy=0,y(1)=1. Then 6y2(e) is equal to

Select Answer:

Visualized Solution

The Given Equation

Rearranging the Terms

Identifying Homogeneity

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

Differentiating the Substitution

Substituting into the Equation

Simplifying the Expression

Separating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Integrating Both Sides

Back-Substitution

Applying the Initial Condition

The Particular Solution

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=e

Final Calculation

Summary and Conclusion

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Magic of Substitution

The Final Integration

Applying Conditions and Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation 2x2dxdy​−2xy+3y2=0 such that y(e)=3e​, then y(1) 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=y(x) be the solution of the differential equation dxdy​=3xy2+x34y3+2yx2​,y(1)=1. If for some n∈N,y(2)∈[n−1,n), then n is equal to

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

The Substitution $y = v x$

Separating $v$ and $x$

Evaluating at $x = e$

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ 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:

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

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

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 :

If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

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

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}, y (1) = 1$ . If for some $n \in N, y (2) \in [n - 1, n)$ , then $n$ is equal to