JEE Main 2023 (25 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ is equal to :

Q: Why can't we just use separation of variables here? It looks like we could move the $y$ terms to one side.

That is a great instinct, but look closer at the term $(1 + xy^2(1 + \log_e x))$. The presence of the $x$ inside the parenthesis, multiplied by $y^2$, couples the variables in a way that makes simple separation impossible. The $y^2$ is trapped with the $x$ terms. This is exactly why we need the Bernoulli transformation—it is the surgical tool required to decouple these variables.

Q: Why did we choose $t = -y^{-2}$? Why not just $t = y^{-2}$?

You absolutely could use $t = y^{-2}$, but using the negative sign is a strategic choice that simplifies the algebra later. When you differentiate $t = -y^{-2}$, you get $\frac{dt}{dx} = 2y^{-3}\frac{dy}{dx}$. This directly matches the term we get after dividing the original equation by $y^3$. It saves you from carrying a negative sign through your integration, reducing the chance of a silly arithmetic error.

Q: I am stuck on the integration part. How do I handle $\int x^2 \log_e x dx$?

When you see a product of an algebraic function ($x^2$) and a logarithmic function ($\log_e x$), the ILATE rule is your best friend. ILATE tells us to choose the logarithmic function as our first function ($u$) and the algebraic function as our second function ($dv$). So, set $u = \log_e x$ and $dv = x^2 dx$. This makes the integration by parts straightforward: $\int u dv = uv - \int v du$.

Select Answer:

Progressive Hints

Hint 1

The Spark: Look closely at the structure of the differential equation. It is not a simple separable equation, nor is it a standard linear one. Notice the

y^{3}

term on the right-hand side. This is a classic signature of a Bernoulli differential equation. Your first task is to rearrange the terms to isolate the non-linear part.

Hint 2

The Arsenal: Once you have identified the Bernoulli form, your toolkit must include the standard substitution method. You need to eliminate the

y^{3}

term. Divide the entire equation by

y^{3}

and then perform a substitution for the term

y^{- 2}

. This will transform the non-linear equation into a linear one, which is much easier to handle.

Hint 3

The Execution: Start by setting

t = - y^{- 2}

. Differentiate this with respect to

x

to find

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

. Substitute these into your equation to get a linear differential equation in terms of

t

and

x

. From there, calculate the Integrating Factor (I.F.) and proceed with the standard linear solution method.

Visualized Solution

Analyze the Differential Equation

Given DE: $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x))$ for $x > 0$
Initial Condition: $y (1) = 3$
Goal: Find the expression for $\frac{y ^{2} ( x )}{9}$

Expand and Rearrange

Expanding the RHS: $\frac{d y}{d x} = \frac{y}{x} + y^{3} (1 + lo g_{e} x)$
Rearranging into Bernoulli form: $\frac{d y}{d x} - \frac{1}{x} y = (1 + lo g_{e} x) y^{3}$

Divide by $y^{3}$

Dividing by $y^{3}$ : $y^{- 3} \frac{d y}{d x} - \frac{1}{x} y^{- 2} = 1 + lo g_{e} x$

Substitution for Linearization

Let $t = - \frac{1}{y ^{2}} = - y^{- 2}$
Differentiating with respect to $x$ : $\frac{d t}{d x} = 2 y^{- 3} \frac{d y}{d x}$
Therefore, $y^{- 3} \frac{d y}{d x} = \frac{1}{2} \frac{d t}{d x}$

Forming the Linear DE

Substituting $t$ and $\frac{d t}{d x}$ : $\frac{1}{2} \frac{d t}{d x} + \frac{t}{x} = 1 + lo g_{e} x$
Standard Linear Form: $\frac{d t}{d x} + \frac{2}{x} t = 2 (1 + lo g_{e} x)$

Calculate Integrating Factor (I.F.)

Integrating Factor $I . F . = e^{\int P d x} = e^{\int \frac{2}{x} d x}$
$I . F . = e^{2 l o g_{e} x} = e^{l o g_{e} x^{2}} = x^{2}$

General Solution Setup

General Solution: $t \cdot (I . F .) = \int Q \cdot (I . F .) d x$
$t \cdot x^{2} = \int 2 (1 + lo g_{e} x) \cdot x^{2} d x$
$t \cdot x^{2} = 2 \int (x^{2} + x^{2} lo g_{e} x) d x$

Integration by Parts

Integrating $x^{2} lo g_{e} x$ by parts ( $u = lo g_{e} x, d v = x^{2} d x$ ):
$\int x^{2} lo g_{e} x d x = \frac{x ^{3}}{3} lo g_{e} x - \int \frac{x ^{3}}{3} \cdot \frac{1}{x} d x$
$= \frac{x ^{3}}{3} lo g_{e} x - \frac{x ^{3}}{9}$

Combine and Simplify

$t \cdot x^{2} = 2 [\frac{x ^{3}}{3} + (\frac{x ^{3}}{3} lo g_{e} x - \frac{x ^{3}}{9})] + C$
$t \cdot x^{2} = \frac{4 x ^{3}}{9} + \frac{2 x ^{3}}{3} lo g_{e} x + C$

Back Substitution

Substitute $t = - \frac{1}{y ^{2}}$ :
$- \frac{x ^{2}}{y ^{2}} = \frac{4 x ^{3}}{9} + \frac{2 x ^{3}}{3} lo g_{e} x + C$

Solve for Constant $C$

At $x = 1, y = 3$ : $- \frac{1 ^{2}}{3 ^{2}} = \frac{4 ( 1 ) ^{3}}{9} + \frac{2 ( 1 ) ^{3}}{3} lo g_{e} 1 + C$
$- \frac{1}{9} = \frac{4}{9} + 0 + C ⟹ C = - \frac{5}{9}$

Final Equation Assembly

$- \frac{x ^{2}}{y ^{2}} = \frac{4 x ^{3} + 6 x ^{3} l o g _{e} x - 5}{9}$
$\frac{x ^{2}}{y ^{2}} = \frac{5 - 4 x ^{3} - 6 x ^{3} l o g _{e} x}{9}$

Rearrange for $\frac{y ^{2}}{9}$

$\frac{y ^{2}}{x ^{2}} = \frac{9}{5 - 2 x ^{3} ( 2 + 3 l o g _{e} x )}$
$\frac{y ^{2}}{9} = \frac{x ^{2}}{5 - 2 x ^{3} ( 2 + 3 l o g _{e} x )}$

Final Result

Using $3 lo g_{e} x = lo g_{e} x^{3}$ :
$\frac{y ^{2}}{9} = \frac{x ^{2}}{5 - 2 x ^{3} ( 2 + l o g _{e} x ^{3} )}$
Correct Option: (0)

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x))

To begin, we expand the right-hand side by distributing the term

\frac{y}{x}

\frac{d y}{d x} = \frac{y}{x} + y^{3} (1 + lo g_{e} x)

Transforming the Equation

The presence of the

y^{3}

term indicates that this is a Bernoulli differential equation. To linearize it, we divide the entire equation by

y^{3}

y^{- 3} \frac{d y}{d x} - \frac{1}{x} y^{- 2} = 1 + lo g_{e} x

We now perform the substitution

t = y^{- 2}

. Differentiating with respect to

x

gives:

\frac{d t}{d x} = - 2 y^{- 3} \frac{d y}{d x} \Rightarrow y^{- 3} \frac{d y}{d x} = - \frac{1}{2} \frac{d t}{d x}

Substituting this into our equation yields:

- \frac{1}{2} \frac{d t}{d x} - \frac{t}{x} = 1 + lo g_{e} x

Multiplying by

- 2

, we obtain the standard linear form:

\frac{d t}{d x} + \frac{2}{x} t = - 2 (1 + lo g_{e} x)

Solving via Integrating Factor

The Integrating Factor (I.F.) is calculated as follows:

I . F . = e^{\int \frac{2}{x} d x} = e^{2 l o g_{e} x} = x^{2}

Multiplying the linear equation by

x^{2}

, we get:

\frac{d}{d x} (t \cdot x^{2}) = - 2 x^{2} (1 + lo g_{e} x)

Integrating both sides with respect to

x

t \cdot x^{2} = - 2 \int (x^{2} + x^{2} lo g_{e} x) d x

Using the ILATE rule for the integral

\int x^{2} lo g_{e} x d x

, we find:

t \cdot x^{2} = - 2 (\frac{x ^{3}}{3} + \frac{x ^{3}}{3} lo g_{e} x - \frac{x ^{3}}{9}) + C

t \cdot x^{2} = - \frac{2 x ^{3}}{9} - \frac{2 x ^{3}}{3} lo g_{e} x + C

Final Calculation

Substituting

t = \frac{1}{y ^{2}}

back into the equation:

\frac{x ^{2}}{y ^{2}} = - \frac{2 x ^{3}}{9} - \frac{2 x ^{3}}{3} lo g_{e} x + C

Using the initial condition

y (1) = 3

, we substitute

x = 1

and

y = 3

\frac{1}{9} = - \frac{2}{9} - 0 + C \Rightarrow C = \frac{3}{9} = \frac{1}{3}

After simplifying the algebraic expression, we arrive at the final result:

\frac{y ^{2}}{9} = \frac{x ^{2}}{3 - 2 x ^{3} ( 1 + lo g _{e} x )}

Common Traps & FAQs

Why can't we just use separation of variables here? It looks like we could move the

y

terms to one side.

That is a great instinct, but look closer at the term

(1 + x y^{2} (1 + lo g_{e} x))

. The presence of the

x

inside the parenthesis, multiplied by

y^{2}

, couples the variables in a way that makes simple separation impossible. The

y^{2}

is trapped with the

x

terms. This is exactly why we need the Bernoulli transformation—it is the surgical tool required to decouple these variables.

Why did we choose

t = - y^{- 2}

? Why not just

t = y^{- 2}

You absolutely could use

t = y^{- 2}

, but using the negative sign is a strategic choice that simplifies the algebra later. When you differentiate

t = - y^{- 2}

, you get

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

. This directly matches the term we get after dividing the original equation by

y^{3}

. It saves you from carrying a negative sign through your integration, reducing the chance of a silly arithmetic error.

I am stuck on the integration part. How do I handle

\int x^{2} lo g_{e} x d x

When you see a product of an algebraic function (

x^{2}

) and a logarithmic function (

lo g_{e} x

), the ILATE rule is your best friend. ILATE tells us to choose the logarithmic function as our first function (

u

) and the algebraic function as our second function (

d v

). So, set

u = lo g_{e} x

and

d v = x^{2} d x

. This makes the integration by parts straightforward:

\int u d v = uv - \int v d u

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Expand and Rearrange

Divide by $y^{3}$

Substitution for Linearization

Forming the Linear DE

Calculate Integrating Factor (I.F.)

General Solution Setup

Integration by Parts

Combine and Simplify

Back Substitution

Solve for Constant $C$

Final Equation Assembly

Rearrange for $\frac{y ^{2}}{9}$

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Transforming the Equation

Solving via Integrating Factor

Final Calculation

Similar Questions

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

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

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x ^{2} - 1} y = (\frac{x - 1}{x + 1})^{1/2}, x > 1$ passing through the point $(2, \frac{1}{3})$ . Then $7 y (8)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$ , which passes through the point $(0, 1)$ . Then $y (1)$ is equal to:

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} + \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{\frac{3}{2}}} y = 2 x exp (\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )})$ pass through the origin. Then $y (1)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

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

If the curve $y = y (x)$ is the solution of the differential equation $2 (x^{2} + x^{5/4}) d y - y (x + x^{1/4}) d x = 2 x^{9/4} d x, x > 0$ which passes through the point $(1, 1 - \frac{4}{3} lo g_{e} 2)$ , then the value of $y (16)$ is equal to :

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

Select Answer:

Visualized Solution

Analyze the Differential Equation

Expand and Rearrange

Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y3

Substitution for Linearization

Forming the Linear DE

Calculate Integrating Factor (I.F.)

General Solution Setup

Integration by Parts

Combine and Simplify

Back Substitution

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

Final Equation Assembly

Rearrange for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9y2​

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Transforming the Equation

Solving via Integrating Factor

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 curve of the differential equation x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(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 curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0 passing through the point (1,0). 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 curve of the differential equation dxdy​+x2−11​y=(x+1x−1​)1/2,x>1 passing through the point (2,3​1​). Then 7​y(8) 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 cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx, 0≤x≤2π​,y(0)=0. Then y(3π​) 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 xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(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 (2xloge​x)dxdy​+2y=x3​loge​x,x>0 and y(e−1)=0. Then, y(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 xdxdy​+2y=x2(x=0) with y(1)=1, is

Divide by $y^{3}$

Solve for Constant $C$

Rearrange for $\frac{y ^{2}}{9}$

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

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

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x ^{2} - 1} y = (\frac{x - 1}{x + 1})^{1/2}, x > 1$ passing through the point $(2, \frac{1}{3})$ . Then $7 y (8)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

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

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