JEE Main 2022 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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

Q: Wait, why did we assume $y = vx$? How do I know when to use that substitution?

Great question! We use $y = vx$ whenever the differential equation is 'homogeneous'. This means every single term in the numerator and denominator has the same total degree. In our case, the numerator terms $4y^3$ and $2yx^2$ are both degree 3, and the denominator terms $3xy^2$ and $x^3$ are also degree 3. This symmetry allows us to cancel out the $x$ variables entirely, turning a scary-looking equation into a simple separable one.

Q: Why does the Intermediate Value Theorem guarantee the root is between 2 and 3?

The function $f(y) = y^3 + 4y - 32$ is a polynomial, which means it is continuous everywhere. If $f(2)$ is negative and $f(3)$ is positive, the graph must cross the x-axis somewhere between $y=2$ and $y=3$ to get from below the axis to above it. That crossing point is our root!

Q: What if I made a mistake in the integration step? How can I check?

Always differentiate your result! If you think $\int \frac{3v^2 + 1}{v^3 + v} dv = \ln|v^3 + v|$, just take the derivative of $\ln|v^3 + v|$ using the chain rule. You will get $\frac{1}{v^3 + v} \cdot (3v^2 + 1)$, which perfectly matches the integrand. It's the best way to build confidence during the exam.

Enter Numerical Value:

Visualized Solution

Identifying the Type of Equation

Given equation: $\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}$
Observe that the degrees of all terms in the numerator and denominator are equal to $3$ .
This is a Homogeneous Differential Equation.

Substitution: $y = v x$

Let $y = v x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d y}{d x} = v + x \frac{d v}{d x}$

Transforming the Equation

Substitute $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ into the DE:
$v + x \frac{d v}{d x} = \frac{4 ( v x ) ^{3} + 2 ( v x ) x ^{2}}{3 x ( v x ) ^{2} + x ^{3}}$
$v + x \frac{d v}{d x} = \frac{4 v ^{3} x ^{3} + 2 v x ^{3}}{3 v ^{2} x ^{3} + x ^{3}}$
Cancel $x^{3}$ from numerator and denominator: $v + x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v}{3 v ^{2} + 1}$

Isolating the $v$ Terms

Isolate $x \frac{d v}{d x}$ :
$x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v}{3 v ^{2} + 1} - v$
$x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v - v ( 3 v ^{2} + 1 )}{3 v ^{2} + 1}$
$x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v - 3 v ^{3} - v}{3 v ^{2} + 1}$
$x \frac{d v}{d x} = \frac{v ^{3} + v}{3 v ^{2} + 1}$

Variable Separation

Rearrange terms to separate $v$ and $x$ :
$\frac{3 v ^{2} + 1}{v ^{3} + v} d v = \frac{1}{x} d x$
Now we are ready to integrate both sides.
$\int \frac{3 v ^{2} + 1}{v ^{3} + v} d v = \int \frac{1}{x} d x$

Integration

Notice that $\frac{d}{d v} (v^{3} + v) = 3 v^{2} + 1$ .
The left side is of the form $\int \frac{f ^{'} ( v )}{f ( v )} d v = ln ∣ f (v) ∣$ .
Integrating both sides: $ln ∣ v^{3} + v ∣ = ln ∣ x ∣ + ln C$
Using logarithm properties: $ln ∣ v^{3} + v ∣ = ln ∣ C x ∣$

General Equation

Remove logarithms: $v^{3} + v = C x$
Substitute $v = \frac{y}{x}$ back into the equation:
$(\frac{y}{x})^{3} + \frac{y}{x} = C x$
Multiply by $x^{3}$ to simplify: $y^{3} + y x^{2} = C x^{4}$

Finding the Constant $C$

Use the initial condition $y (1) = 1$ :
$1^{3} + 1 (1)^{2} = C (1)^{4}$
$1 + 1 = C ⟹ C = 2$
The particular solution is: $y^{3} + y x^{2} = 2 x^{4}$

Evaluating at $x = 2$

Substitute $x = 2$ into the equation to find $y (2)$ :
$y^{3} + y (2)^{2} = 2 (2)^{4}$
$y^{3} + 4 y = 2 (16)$
$y^{3} + 4 y - 32 = 0$

Estimating the Root at $y = 2$

Let $f (y) = y^{3} + 4 y - 32$ .
Check $y = 2$ : $f (2) = 2^{3} + 4 (2) - 32$
$f (2) = 8 + 8 - 32 = - 16$
Since $- 16 < 0$ , the function is negative at $y = 2$ .

Estimating the Root at $y = 3$

Check $y = 3$ : $f (3) = 3^{3} + 4 (3) - 32$
$f (3) = 27 + 12 - 32 = 7$
Since $7 > 0$ , the function is positive at $y = 3$ .
Because $f (y)$ is continuous and changes sign, the root $y (2) \in (2, 3)$ .

Final Conclusion

Given condition: $y (2) \in [n - 1, n)$ .
We found: $y (2) \in (2, 3)$ .
Comparing the intervals, we get $n - 1 = 2$ and $n = 3$ .
Therefore, $n = 3$ .

00:00 / 00:00

The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving a differential equation; we are uncovering a hidden symmetry.

When you first look at the equation

\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}

, it might seem like a chaotic mess of variables. But look closer at the powers.

The numerator has terms of degree 3 (

y^{3}

and

y x^{2}

), and the denominator also has terms of degree 3 (

x y^{2}

and

x^{3}

). This is a Homogeneous Differential Equation, and it is begging to be simplified.

The Transformation

Entering the

v

-Space

To tame this beast, we use the classic substitution

y = v x

. By doing this, we are essentially changing our perspective, looking at the ratio of

y

x

rather than the variables themselves.

When we differentiate

y = v x

with respect to

x

, we apply the product rule to get

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

. Now, substitute this into our original equation:

v + x \frac{d v}{d x} = \frac{4 ( v x ) ^{3} + 2 ( v x ) x ^{2}}{3 x ( v x ) ^{2} + x ^{3}}

Watch the magic happen. Every single term contains an

x^{3}

. We can factor it out and cancel it completely, leaving us with a beautiful, clean expression:

v + x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v}{3 v ^{2} + 1}

The

x

dependence is now isolated, and we are ready to move forward.

The Dance of Integration

We isolate the differential terms to get

x \frac{d v}{d x} = \frac{4 v ^{3} + 2 v}{3 v ^{2} + 1} - v

. Simplifying this, we arrive at:

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

Now, we separate the variables:

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

Here is where the elegance of calculus shines. The numerator

3 v^{2} + 1

is exactly the derivative of the denominator

v^{3} + v

. This is the golden form

\int \frac{f ^{'} ( v )}{f ( v )} d v = ln ∣ f (v) ∣

Integrating both sides gives us

ln ∣ v^{3} + v ∣ = ln ∣ x ∣ + ln C

, which simplifies to

v^{3} + v = C x

. Replacing

v

with

\frac{y}{x}

, we get the general solution:

y^{3} + y x^{2} = C x^{4}

The Final Reveal

We are given the initial condition

y (1) = 1

. Plugging these values in, we find

1^{3} + 1 (1)^{2} = C (1)^{4}

, which gives us $C = 2$ .

Our particular solution is

y^{3} + y x^{2} = 2 x^{4}

. Now, we need to find

y (2)

. Setting

x = 2

, we get

y^{3} + 4 y = 2 (2^{4})

, or:

y^{3} + 4 y - 32 = 0

We don't need to solve this cubic exactly. We just need to trap it. Let

f (y) = y^{3} + 4 y - 32

- At

y = 2

f (2) = 8 + 8 - 32 = - 16

. - At

y = 3

f (3) = 27 + 12 - 32 = 7

Since the function changes sign between

y = 2

and

y = 3

, the root must lie in the interval

(2, 3)

. Given the condition $y(2) \in

Common Traps & FAQs

Wait, why did we assume

y = v x

? How do I know when to use that substitution?

Great question! We use

y = v x

whenever the differential equation is 'homogeneous'. This means every single term in the numerator and denominator has the same total degree. In our case, the numerator terms

4 y^{3}

and

2 y x^{2}

are both degree 3, and the denominator terms

3 x y^{2}

and

x^{3}

are also degree 3. This symmetry allows us to cancel out the

x

variables entirely, turning a scary-looking equation into a simple separable one.

Why did we get

ln ∣ v^{3} + v ∣ = ln ∣ C x ∣

? Can't we just drop the absolute value bars?

In a formal sense, we keep the absolute value bars to maintain mathematical rigor. However, since we are dealing with a physical trajectory starting at

y (1) = 1

, the values of

v

and

x

remain positive in the region of interest. Thus, the expression inside the logarithm remains positive, and we can safely work with the algebraic form

v^{3} + v = C x

Is there a faster way to solve

y^{3} + 4 y - 32 = 0

without guessing values?

For a cubic equation, you could use Cardano's method, but that is overkill for a JEE exam! The question asks for an interval

[n - 1, n)

, which is a massive hint that we don't need the exact root. We only need to prove the root lies between two integers. Testing

y = 2

and

y = 3

is the most efficient 'trap-catcher' strategy.

Why does the Intermediate Value Theorem guarantee the root is between 2 and 3?

The function

f (y) = y^{3} + 4 y - 32

is a polynomial, which means it is continuous everywhere. If

f (2)

is negative and

f (3)

is positive, the graph must cross the x-axis somewhere between

y = 2

and

y = 3

to get from below the axis to above it. That crossing point is our root!

What if I made a mistake in the integration step? How can I check?

Always differentiate your result! If you think

\int \frac{3 v ^{2} + 1}{v ^{3} + v} d v = ln ∣ v^{3} + v ∣

, just take the derivative of

ln ∣ v^{3} + v ∣

using the chain rule. You will get

\frac{1}{v ^{3} + v} \cdot (3 v^{2} + 1)

, which perfectly matches the integrand. It's the best way to build confidence during the exam.

Animated Solution for Mathematics - Differential Equations: 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

Enter Numerical Value:

Visualized Solution

Identifying the Type of Equation

Substitution: $y = v x$

Transforming the Equation

Isolating the $v$ Terms

Variable Separation

Integration

General Equation

Finding the Constant $C$

Evaluating at $x = 2$

Estimating the Root at $y = 2$

Estimating the Root at $y = 3$

Final Conclusion

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Dance of Integration

The Final Reveal

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 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

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 :

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

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 $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ 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 $y = y (x)$ 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

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

If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :

Enter Numerical Value:

Visualized Solution

Identifying the Type of Equation

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

Transforming the Equation

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

Variable Separation

Integration

General Equation

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

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Estimating the Root at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2

Estimating the Root at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=3

Final Conclusion

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Dance of Integration

The Final Reveal

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

.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; } 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; } 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; } If sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos(xy​)dxdy​=ycos(xy​)+x and y(1)=3π​, then α2 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 (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; } 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 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; } If ydxdy​=x[x2y2​+ϕ′(x2y2​)ϕ(x2y2​)​],x>0,ϕ>0, and y(1)=−1, then ϕ(4y2​) is equal to :

Substitution: $y = v x$

Isolating the $v$ Terms

Finding the Constant $C$

Evaluating at $x = 2$

Estimating the Root at $y = 2$

Estimating the Root at $y = 3$

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 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

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 :

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

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 $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ 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 $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :