JEE Main 2022 (25 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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

Q: Why do we use the natural log during integration? Is it always $\ln|x|$?

Yes, it is! Whenever you integrate $\frac{1}{x} dx$, the result is $\ln|x| + C$. The absolute value bars are crucial because the domain of the natural logarithm is only positive numbers. In JEE problems, always keep the absolute value bars in mind, though often the context of the problem (like $x=e$) will imply positive values, allowing you to drop them later.

Select Answer:

Visualized Solution

$2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$

Given equation: $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$

$Homogeneous Check$

Degrees of $x^{2}$ , $x y$ , and $y^{2}$ are all $2$ .
Thus, it is a Homogeneous Differential Equation.

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

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

$Substitution y = v x$

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

$Transforming the Equation$

Substitute $\frac{d y}{d x}$ and $\frac{y}{x}$ :
$v + x \frac{d v}{d x} = v - \frac{3}{2} v^{2}$

$Simplifying$

Cancel $v$ from both sides:
$x \frac{d v}{d x} = - \frac{3}{2} v^{2}$

$Variable Separation$

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

$Integration$

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

$Back Substitution$

Substitute $v = \frac{y}{x}$ :
$- \frac{x}{y} = - \frac{3}{2} ln ∣ x ∣ + C$

$Applying Initial Condition$

Given $y (e) = \frac{e}{3}$ .
Substitute $x = e$ and $y = \frac{e}{3}$ :
$- \frac{e}{\frac{e}{3}} = - \frac{3}{2} ln ∣ e ∣ + C$

$Finding C$

$- 3 = - \frac{3}{2} (1) + C$
$C = - 3 + \frac{3}{2} = - \frac{3}{2}$

$Equation for y (1)$

Substitute $C = - \frac{3}{2}$ and $x = 1$ :
$- \frac{1}{y} = - \frac{3}{2} ln ∣1∣ - \frac{3}{2}$

$Final Answer$

Since $ln 1 = 0$ :
$- \frac{1}{y} = - \frac{3}{2}$
$y = \frac{2}{3}$

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The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The given differential equation is

2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0

Upon inspection, we observe that every term—

x^{2}

x y

, and

y^{2}

—possesses a degree of 2. This confirms that the expression is a Homogeneous Differential Equation, implying the system scales uniformly.

The Transformation

To solve this, we apply the substitution

y = v x

, where

v = \frac{y}{x}

. Consequently, by the product rule, the derivative becomes:

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

Substituting these into the original equation, we obtain:

2 x^{2} (v + x \frac{d v}{d x}) - 2 x (v x) + 3 (v x)^{2} = 0

After algebraic simplification, the terms involving

v

cancel out, reducing the equation to:

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

The Calculus of Separation

We now separate the variables to isolate

v

and

x

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

Integrating both sides yields:

- \frac{1}{v} = - \frac{3}{2} ln ∣ x ∣ + C

The Final Reveal

Substituting

v = \frac{y}{x}

back into the equation, we get:

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

Using the initial condition

y (e) = \frac{e}{3}

, we substitute

x = e

and

y = \frac{e}{3}

to determine the constant

C

- \frac{e}{e /3} = - \frac{3}{2} ln (e) + C \Rightarrow - 3 = - \frac{3}{2} + C \Rightarrow C = - \frac{3}{2}

Finally, to find

y (1)

, we set

x = 1

in the equation

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

- \frac{1}{y} = - \frac{3}{2} ln (1) - \frac{3}{2}

Since

ln (1) = 0

, the equation simplifies to

- \frac{1}{y} = - \frac{3}{2}

. Solving for

y

, we arrive at the final answer:

$y = \frac{2}{3}$

Common Traps & FAQs

Wait, how do I know for sure this is a homogeneous equation? Do I just look at the powers?

Exactly! You are looking for the 'degree' of each term. In the equation

2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0

, the term

x^{2}

has degree 2. The term

x y

has degree

1 + 1 = 2

. The term

y^{2}

has degree 2. Because every term has the same degree, the equation is homogeneous. If even one term had a different degree, this method would fail, and you would need to look for a different approach, like a linear differential equation or Bernoulli's form.

Why does the

v

term cancel out during the simplification step? Did I make a mistake?

You absolutely did not make a mistake! In fact, that cancellation is the 'magic' of the homogeneous substitution method. It is designed to happen. When you substitute

y = v x

into a homogeneous equation, the

v

terms will always align in a way that allows you to isolate the

\frac{d v}{d x}

term. If they didn't cancel, it would be a sign that the equation wasn't truly homogeneous or that a calculation error occurred earlier.

Why do we use the natural log during integration? Is it always

ln ∣ x ∣

Yes, it is! Whenever you integrate

\frac{1}{x} d x

, the result is

ln ∣ x ∣ + C

. The absolute value bars are crucial because the domain of the natural logarithm is only positive numbers. In JEE problems, always keep the absolute value bars in mind, though often the context of the problem (like

x = e

) will imply positive values, allowing you to drop them later.

Animated Solution for Mathematics - Differential Equations: 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

Select Answer:

Visualized Solution

$2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$

$Homogeneous Check$

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

$Substitution y = v x$

$Transforming the Equation$

$Simplifying$

$Variable Separation$

$Integration$

$Back Substitution$

$Applying Initial Condition$

$Finding C$

$Equation for y (1)$

$Final Answer$

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Calculus of Separation

The Final Reveal

Similar Questions

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

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 :

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

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 $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ 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 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x2dxdy​−2xy+3y2=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Homogeneous Check

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Transforming the Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Variable Separation

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for y(1)

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

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Calculus of Separation

The Final Reveal

Similar Questions

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=−(3x2+y2x2+3y2​),y(1)=0 is

.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; } If dxdy​=x2+y2xy​; y(1)=1: then a value of x satisfying y(x)=e is:

.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; } 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=xx+y​ satisfying the condition y(1)=1 is

.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 xdxdy​−y=y2+16x2​,y(1)=3 be y=y(x). Then y(2) is equal to :

.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; } 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

$2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$

$Homogeneous Check$

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

$Substitution y = v x$

$Transforming the Equation$

$Simplifying$

$Variable Separation$

$Integration$

$Back Substitution$

$Applying Initial Condition$

$Finding C$

$Equation for y (1)$

$Final Answer$

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

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 :

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

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 $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ 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 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