JEE Main 2023 (24 January Shift 1)

LEVELJEE Main

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

Q: Why do we divide by $x^3 dx$ instead of just $dx$?

Great question! If you only divide by $dx$, you get $x^3 \frac{dy}{dx} + xy - 1 = 0$. While this is mathematically correct, it doesn't immediately reveal the linear form. By dividing by $x^3$, we clear the coefficient of $\frac{dy}{dx}$, making it a monic linear differential equation. This is the standard form required to apply the Integrating Factor method effectively.

Q: Do I really need the constant $C$?

Absolutely! Without $C$, you have a general solution—a family of curves. The initial condition $y(\frac{1}{2}) = 3 - e$ is the specific key that unlocks the one curve passing through that exact point. Forgetting $C$ is like having a map but not knowing where you are standing on it.

Select Answer:

Progressive Hints

Hint 1

Don't let the

x^{3}

term intimidate you. Your first objective is to isolate the derivative

\frac{d y}{d x}

. Once you divide the entire equation by

x^{3} d x

, look closely at the structure. Does it resemble the standard linear form

\frac{d y}{d x} + P (x) y = Q (x)

? That is your gateway to the solution.

Hint 2

Once you have identified

P (x) = \frac{1}{x ^{2}}

, the path forward is paved by the Integrating Factor (I.F.). Remember, the I.F. is defined as

e^{\int P (x) d x}

. Calculate this integral carefully; the negative sign in the exponent is crucial for the next step.

Hint 3

After multiplying by the I.F., you will face an integral of the form

\int \frac{1}{x ^{3}} e^{- 1/ x} d x

. This looks daunting, but use the substitution

t = - \frac{1}{x}

. This transforms the integral into a standard integration by parts problem involving

t e^{t}

. Watch how the

x^{3}

term breaks down into

t

and

d t

Visualized Solution

The Given Equation

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

Rearranging the Equation

Divide the entire equation by $x^{3} d x$ :
$\frac{d y}{d x} + \frac{x y - 1}{x ^{3}} = 0$

Standard Linear Form

Separate the terms in the fraction:
$\frac{d y}{d x} + \frac{y}{x ^{2}} - \frac{1}{x ^{3}} = 0$
Standard form: $\frac{d y}{d x} + P (x) y = Q (x)$

Identifying $P (x)$ and $Q (x)$

Comparing with $\frac{d y}{d x} + P (x) y = Q (x)$ :
$P (x) = \frac{1}{x ^{2}}$
$Q (x) = \frac{1}{x ^{3}}$

Integrating Factor (I.F.)

Formula: $I . F . = e^{\int P (x) d x}$
Substitute $P (x)$ :
$I . F . = e^{\int \frac{1}{x ^{2}} d x}$

Calculating I.F.

Evaluate the integral:
$\int \frac{1}{x ^{2}} d x = - \frac{1}{x}$
Therefore, $I . F . = e^{- \frac{1}{x}}$

General Solution Setup

Formula: $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x$
Substitute values:
$y \cdot e^{- \frac{1}{x}} = \int \frac{1}{x ^{3}} e^{- \frac{1}{x}} d x$

Substitution for Integration

Let $- \frac{1}{x} = t$
Differentiating: $\frac{1}{x ^{2}} d x = d t$
Also, $\frac{1}{x} = - t$

Transforming the Integral

Rewrite $\frac{1}{x ^{3}}$ as $\frac{1}{x} \cdot \frac{1}{x ^{2}}$
Integral becomes: $\int (- t) e^{t} d t$
$= - \int t e^{t} d t$

Integration by Parts

Using $\int u d v = uv - \int v d u$
$- \int t e^{t} d t = - (t e^{t} - \int e^{t} d t)$
$= - (t e^{t} - e^{t}) + C$
$= e^{t} (1 - t) + C$

General Solution for $y$

Substitute $t = - \frac{1}{x}$ back:
$y \cdot e^{- \frac{1}{x}} = e^{- \frac{1}{x}} (1 + \frac{1}{x}) + C$
Divide by $e^{- \frac{1}{x}}$ :
$y (x) = 1 + \frac{1}{x} + C e^{\frac{1}{x}}$

Using Initial Condition

Given: $y (\frac{1}{2}) = 3 - e$
Substitute $x = \frac{1}{2}$ and $y = 3 - e$ :
$3 - e = 1 + 2 + C e^{2}$

Finding the Constant $C$

Simplify: $3 - e = 3 + C e^{2}$
Cancel $3$ from both sides:
$- e = C e^{2}$
$C = - \frac{1}{e}$

Finding $y (1)$

Specific solution: $y (x) = 1 + \frac{1}{x} - \frac{1}{e} e^{\frac{1}{x}}$
Substitute $x = 1$ :
$y (1) = 1 + 1 - \frac{1}{e} e^{1}$
$y (1) = 2 - 1 = 1$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We begin with the differential equation:

x^{3} d y + (x y - 1) d x = 0

To reveal the underlying structure, we isolate the derivative

\frac{d y}{d x}

. Dividing the entire equation by

x^{3} d x

, we obtain:

\frac{d y}{d x} + \frac{x y - 1}{x ^{3}} = 0

By splitting the fraction, we rewrite the equation as:

\frac{d y}{d x} + \frac{y}{x ^{2}} - \frac{1}{x ^{3}} = 0

The Master Equation

Moving the constant term to the right side, we arrive at the standard form of a Linear Differential Equation:

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

This matches the form

\frac{d y}{d x} + P (x) y = Q (x)

, where

P (x) = \frac{1}{x ^{2}}

and

Q (x) = \frac{1}{x ^{3}}

Calculating the Integrating Factor

We now determine the Integrating Factor (I.F.), defined as

e^{\int P (x) d x}

I . F . = e^{\int \frac{1}{x ^{2}} d x} = e^{- 1/ x}

Multiplying the entire differential equation by this I.F. allows the left side to collapse into the derivative of a product:

\frac{d}{d x} [y \cdot e^{- 1/ x}] = \frac{1}{x ^{3}} e^{- 1/ x}

Solving the Integral

To solve for

y

, we integrate both sides:

y \cdot e^{- 1/ x} = \int \frac{1}{x ^{3}} e^{- 1/ x} d x

Using the substitution

t = - \frac{1}{x}

, we have

d t = \frac{1}{x ^{2}} d x

. The integral transforms into:

\int (- t) e^{t} d t

Applying integration by parts (

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

) with

u = t

and

d v = e^{t} d t

, we get:

- (t e^{t} - e^{t}) + C = e^{t} (1 - t) + C

Substituting

t = - \frac{1}{x}

back into the expression, we find:

y \cdot e^{- 1/ x} = e^{- 1/ x} (1 + \frac{1}{x}) + C

Dividing by

e^{- 1/ x}

, the general solution is:

y (x) = 1 + \frac{1}{x} + C e^{1/ x}

Final Calculation

We apply the initial condition

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

to determine the constant

C

3 - e = 1 + \frac{1}{1/2} + C e^{1/ (1/2)}

3 - e = 1 + 2 + C e^{2}

- e = C e^{2} \Rightarrow C = - \frac{1}{e}

The specific solution is:

y (x) = 1 + \frac{1}{x} - \frac{1}{e} e^{1/ x}

Evaluating at

x = 1

y (1) = 1 + 1 - \frac{1}{e} e^{1} = 2 - 1 = 1

The final answer is 1.

Common Traps & FAQs

Why do we divide by

x^{3} d x

instead of just

d x

Great question! If you only divide by

d x

, you get

x^{3} \frac{d y}{d x} + x y - 1 = 0

. While this is mathematically correct, it doesn't immediately reveal the linear form. By dividing by

x^{3}

, we clear the coefficient of

\frac{d y}{d x}

, making it a monic linear differential equation. This is the standard form required to apply the Integrating Factor method effectively.

I got stuck at the integration step. Why is the substitution

t = - \frac{1}{x}

so effective?

It is the 'Aha!' moment of the problem. Notice that the derivative of

- \frac{1}{x}

\frac{1}{x ^{2}}

. Since we have a

\frac{1}{x ^{3}}

term, we can split it into

\frac{1}{x} \cdot \frac{1}{x ^{2}}

. The

\frac{1}{x ^{2}} d x

becomes

d t

, and the

\frac{1}{x}

becomes

- t

. It turns a complex exponential integral into a simple polynomial-exponential product!

Do I really need the constant

C

Absolutely! Without

C

, you have a general solution—a family of curves. The initial condition

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

is the specific key that unlocks the one curve passing through that exact point. Forgetting

C

is like having a map but not knowing where you are standing on it.

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

Select Answer:

Visualized Solution

The Given Equation

Rearranging the Equation

Standard Linear Form

Identifying $P (x)$ and $Q (x)$

Integrating Factor (I.F.)

Calculating I.F.

General Solution Setup

Substitution for Integration

Transforming the Integral

Integration by Parts

General Solution for $y$

Using Initial Condition

Finding the Constant $C$

Finding $y (1)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Calculating the Integrating Factor

Solving the Integral

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

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

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ is equal to :

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ 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 $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to

Let $y (x)$ be the solution of the differential equation $(x lo g x) \frac{d y}{d x} + y = 2 x lo g x, (x \geq 1)$ . Then $y (e)$ is equal to:

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

Select Answer:

Visualized Solution

The Given Equation

Rearranging the Equation

Standard Linear Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Integrating Factor (I.F.)

Calculating I.F.

General Solution Setup

Substitution for Integration

Transforming the Integral

Integration by Parts

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

Using Initial Condition

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

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

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Calculating the Integrating Factor

Solving the Integral

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=f(x) is the solution of the differential equation ey(dxdy​−1)=ex such that y(0)=0, 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 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; } If x=x(y) is the solution of the differential equation ydydx​=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=0,x>0. If y(e)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0,y=y(x) is the solution of the differential equation (1+x)dy=[(1+x)2+y−3]dx,y(2)=0, 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 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; } If y(t) is a solution of (1+t)dtdy​−ty=1 and y(0)=−1, 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 y(x) be the solution of the differential equation (xlogx)dxdy​+y=2xlogx,(x≥1). Then y(e) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0, y=y(x) is the solution of the differential equation, (1+x)dxdy​=((1+x)2+(y−3)), If y(2)=0 then y(3)=?

Identifying $P (x)$ and $Q (x)$

General Solution for $y$

Finding the Constant $C$

Finding $y (1)$

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

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

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ is equal to :

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ 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 $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to

Let $y (x)$ be the solution of the differential equation $(x lo g x) \frac{d y}{d x} + y = 2 x lo g x, (x \geq 1)$ . Then $y (e)$ is equal to:

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