JEE Main 2024 (06 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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

Q: Why do we divide by $2x \log_e x$ at the start?

Great question! We divide by $2x \log_e x$ to bring the equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. This form is essential because it allows us to use the Integrating Factor method, which is the most powerful tool for solving first-order linear differential equations.

Q: What happens to the constant $C$ when we use the initial condition?

The initial condition $y(e^{-1}) = 0$ is like a GPS coordinate for our solution curve. By plugging in $x = e^{-1}$ and $y = 0$, we solve for the specific value of $C$ that makes the curve pass through that exact point. In this problem, the terms cancel out beautifully to give $C = 0$.

Select Answer:

Visualized Solution

The Differential Equation

$(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x$
Domain: $x > 0$

Standard Linear Form

Divide by $2 x lo g_{e} x$
$\frac{d y}{d x} + (\frac{1}{x l o g _{e} x}) y = \frac{3}{2 x ^{2}}$
Form: $\frac{d y}{d x} + P (x) y = Q (x)$

Integrating Factor (IF)

$I F = e^{\int P (x) d x}$
$I F = e^{\int \frac{1}{x l o g _{e} x} d x}$

Calculating IF

Let $u = lo g_{e} x ⟹ d u = \frac{1}{x} d x$
$\int \frac{1}{u} d u = lo g_{e} u = lo g_{e} (lo g_{e} x)$
$I F = e^{l o g_{e} (l o g_{e} x)} = lo g_{e} x$

General Solution Equation

$y \cdot I F = \int Q (x) \cdot I F d x + C$
$y lo g_{e} x = \int (\frac{3}{2 x ^{2}}) lo g_{e} x d x + C$

Integration by Parts

$\int x^{- 2} lo g_{e} x d x$
Using ILATE rule:
$u = lo g_{e} x$ (Logarithmic)
$d v = x^{- 2} d x$ (Algebraic)

Executing the Integral

$\int u d v = uv - \int v d u$
$v = \int x^{- 2} d x = - \frac{1}{x}$
$= (lo g_{e} x) (- \frac{1}{x}) - \int (- \frac{1}{x}) \frac{1}{x} d x$

Completing the Integration

$= - \frac{l o g _{e} x}{x} + \int x^{- 2} d x$
$= - \frac{l o g _{e} x}{x} - \frac{1}{x}$
$= - \frac{l o g _{e} x + 1}{x}$

The General Solution

$y lo g_{e} x = \frac{3}{2} [- \frac{l o g _{e} x + 1}{x}] + C$
$y lo g_{e} x = - \frac{3 ( l o g _{e} x + 1 )}{2 x} + C$

Finding the Constant $C$

Given: $y (e^{- 1}) = 0$
Substitute $x = e^{- 1}, y = 0$
$0 \cdot lo g_{e} (e^{- 1}) = - \frac{3 ( l o g _{e} ( e ^{- 1} ) + 1 )}{2 e ^{- 1}} + C$

Evaluating $C$

$lo g_{e} (e^{- 1}) = - 1$
$0 = - \frac{3 ( - 1 + 1 )}{2 e ^{- 1}} + C$
$0 = 0 + C ⟹ C = 0$

The Particular Solution

$y lo g_{e} x = - \frac{3 ( l o g _{e} x + 1 )}{2 x}$
$y = - \frac{3 ( l o g _{e} x + 1 )}{2 x l o g _{e} x}$

Finding $y (e)$

Substitute $x = e$
$y (e) \cdot lo g_{e} e = - \frac{3 ( l o g _{e} e + 1 )}{2 e}$
$y (e) \cdot 1 = - \frac{3 ( 1 + 1 )}{2 e}$

Final Answer

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

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x

To solve this, we first standardize the equation into the linear form

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

. Dividing the entire equation by

(2 x lo g_{e} x)

, we obtain:

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

From this, we identify the components:

P (x) = \frac{1}{x l o g _{e} x}

and

Q (x) = \frac{3}{2 x ^{2}}

Finding the Integrating Factor

The Integrating Factor (IF) is defined as

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

. Substituting our

P (x)

I F = e^{\int \frac{1}{x l o g _{e} x} d x}

Using the substitution

u = lo g_{e} x

, where

d u = \frac{1}{x} d x

, the integral becomes:

\int \frac{1}{u} d u = lo g_{e} (lo g_{e} x)

Thus, the Integrating Factor simplifies to:

I F = e^{l o g_{e} (l o g_{e} x)} = lo g_{e} x

The Master Equation

The general solution is given by the formula

y \cdot I F = \int Q (x) \cdot I F d x + C

. Substituting our known values:

y lo g_{e} x = \int \frac{3}{2 x ^{2}} lo g_{e} x d x + C

To solve the integral

\int \frac{3}{2 x ^{2}} lo g_{e} x d x

, we apply Integration by Parts using the ILATE rule. Let

u = lo g_{e} x

and

d v = x^{- 2} d x

Applying

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

, we find:

\int \frac{3}{2 x ^{2}} lo g_{e} x d x = \frac{3}{2} (- \frac{lo g _{e} x}{x} - \int - \frac{1}{x ^{2}} d x) = - \frac{3 ( lo g _{e} x + 1 )}{2 x}

Final Calculation

Combining these, the general solution is:

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

We use the condition

y (e^{- 1}) = 0

to determine the constant

C

. Substituting

x = e^{- 1}

and

y = 0

0 \cdot lo g_{e} (e^{- 1}) = - \frac{3 ( lo g _{e} ( e ^{- 1} ) + 1 )}{2 e ^{- 1}} + C \Rightarrow 0 = - \frac{3 ( - 1 + 1 )}{2 e ^{- 1}} + C \Rightarrow C = 0

With

C = 0

, the particular solution is:

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

Finally, evaluating at

x = e

y (e) = - \frac{3 ( lo g _{e} e + 1 )}{2 e lo g _{e} e} = - \frac{3 ( 1 + 1 )}{2 e ( 1 )} = - \frac{6}{2 e} = - \frac{3}{e}

The final result is $- \frac{3}{e}$ .

Common Traps & FAQs

Why do we divide by

2 x lo g_{e} x

at the start?

Great question! We divide by

2 x lo g_{e} x

to bring the equation into the standard linear form

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

. This form is essential because it allows us to use the Integrating Factor method, which is the most powerful tool for solving first-order linear differential equations.

How do I know when to use Integration by Parts?

Whenever you see an integral that is a product of two different types of functions—like an algebraic function (

x^{- 2}

) and a logarithmic function (

lo g_{e} x

)—Integration by Parts is your best friend. Remember the ILATE rule to choose your

u

and

d v

wisely!

What happens to the constant

C

when we use the initial condition?

The initial condition

y (e^{- 1}) = 0

is like a GPS coordinate for our solution curve. By plugging in

x = e^{- 1}

and

y = 0

, we solve for the specific value of

C

that makes the curve pass through that exact point. In this problem, the terms cancel out beautifully to give

C = 0

Animated Solution for Mathematics - Differential Equations: 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

Select Answer:

Visualized Solution

The Differential Equation

Standard Linear Form

Integrating Factor (IF)

Calculating IF

General Solution Equation

Integration by Parts

Executing the Integral

Completing the Integration

The General Solution

Finding the Constant $C$

Evaluating $C$

The Particular Solution

Finding $y (e)$

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Finding the Integrating Factor

The Master Equation

Final Calculation

Similar Questions

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:

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

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 :

Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x lo g_{e} x, (x > 1)$ . If $2 y (2) = lo g_{e} 4 - 1$ , then $y (e)$ is equal to :-

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

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

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 $y = y (x)$ 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 :

Select Answer:

Visualized Solution

The Differential Equation

Standard Linear Form

Integrating Factor (IF)

Calculating IF

General Solution Equation

Integration by Parts

Executing the Integral

Completing the Integration

The General Solution

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

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

The Particular Solution

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

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Finding the Integrating Factor

The Master Equation

Final Calculation

Similar Questions

.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; } 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; } 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; } Let y=y(x) be the solution of the differential equation, xdxdy​+y=xloge​x,(x>1). If 2y(2)=loge​4−1, then y(e) is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(21​)=3−e. 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 of the differential equation dxdy​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.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 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 :

Finding the Constant $C$

Evaluating $C$

Finding $y (e)$

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:

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

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 :

Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x lo g_{e} x, (x > 1)$ . If $2 y (2) = lo g_{e} 4 - 1$ , then $y (e)$ is equal to :-

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

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

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 $y = y (x)$ 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 :