Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = (e^{-2\sqrt{x}}g(f(f(x))) - \frac{y}{\sqrt{x}}), y(0) = 0$, then $y(1)$ is :- Options: (A) $\frac{1-e^2}{e^4}$ (B) $\frac{2e-1}{e^3}$ (C) $\frac{e-1}{e^4}$ (D) $\frac{1-e^3}{e^4}$

### Analyzing the Setup The problem presents a nested function structure that initially appears complex. We are given $f(x) = x - 1$ and $g(x) = e^x$. To simplify the term $g(f(f(x)))$, we first evaluate the inner composition: $$f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2$$ Applying the function $g$ to this result, we obtain: $$g(f(f(x))) = e^{x - 2}$$ ### The Master Equation We now substitute this into the given differential equation: $$\frac{dy}{dx} = e^{-2\sqrt{x}} \cdot e^{x - 2} - \frac{y}{\sqrt{x}}$$ Rearranging the terms to isolate the linear differential form $\frac{dy}{dx} + P(x)y = Q(x)$, we get: $$\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = e^{x - 2\sqrt{x} - 2}$$ Here, we identify the components: $$P(x) = \frac{1}{\sqrt{x}}, \quad Q(x) = e^{x - 2\sqrt{x} - 2}$$ ### The Integrating Factor The Integrating Factor ($IF$) is calculated as follows: $$IF = e^{\int P(x) dx} = e^{\int x^{-1/2} dx} = e^{2\sqrt{x}}$$ Multiplying the entire differential equation by the $IF$, the left side simplifies to the derivative of the product $y \cdot e^{2\sqrt{x}}$: $$\frac{d}{dx} \left( y \cdot e^{2\sqrt{x}} \right) = e^{x - 2\sqrt{x} - 2} \cdot e^{2\sqrt{x}}$$ Simplifying the right side by adding the exponents: $$\frac{d}{dx} \left( y \cdot e^{2\sqrt{x}} \right) = e^{x - 2}$$ ### Final Calculation Integrating both sides with respect to $x$: $$y \cdot e^{2\sqrt{x}} = \int e^{x - 2} dx = e^{x - 2} + C$$ Using the initial condition $y(0) = 0$: $$0 \cdot e^0 = e^{0 - 2} + C \implies 0 = e^{-2} + C \implies C = -e^{-2}$$ The general solution is therefore: $$y \cdot e^{2\sqrt{x}} = e^{x - 2} - e^{-2}$$ To find $y(1)$, we substitute $x = 1$: $$y(1) \cdot e^2 = e^{1 - 2} - e^{-2} = e^{-1} - e^{-2}$$ Dividing by $e^2$, we reach the final result: $$y(1) = \frac{e^{-1} - e^{-2}}{e^2} = \frac{\frac{1}{e} - \frac{1}{e^2}}{e^2} = \frac{e - 1}{e^4}$$ The final answer is **$\frac{e - 1}{e^4}$**.

JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x}), y (0) = 0$ , then $y (1)$ is :-

Q: Is the constant $C$ always zero?

No, never assume $C=0$! Always use the initial condition provided in the problem. Here, we were given $y(0)=0$. We substitute $x=0$ and $y=0$ into our general solution to find the specific value of $C$. In this problem, it turned out to be $-e^{-2}$. Always respect the initial condition; it is the key that unlocks the specific solution from the general family of curves.

Select Answer:

Visualized Solution

Problem Setup

Given: $f (x) = x - 1$ and $g (x) = e^{x}$
Differential Equation: $\frac{d y}{d x} = e^{- 2 x} g (f (f (x))) - \frac{y}{x}$
Initial Condition: $y (0) = 0$
Goal: Find $y (1)$

Evaluating $f (f (x))$

Start with the inner function: $f (x) = x - 1$
Substitute $f (x)$ into itself: $f (f (x)) = f (x - 1)$
Apply the definition of $f$ : $(x - 1) - 1$
Result: $f (f (x)) = x - 2$

Evaluating $g (f (f (x)))$

We have $g (x) = e^{x}$
Substitute our result for $f (f (x))$ : $g (x - 2)$
Apply the definition of $g$ : $e^{x - 2}$
Result: $g (f (f (x))) = e^{x - 2}$

Updating the Differential Equation

Original: $\frac{d y}{d x} = e^{- 2 x} g (f (f (x))) - \frac{y}{x}$
Substitute the composite function: $\frac{d y}{d x} = e^{- 2 x} \cdot e^{x - 2} - \frac{y}{x}$
Combine the exponents on the right: $e^{x - 2 x - 2}$

Standard Linear Form

Move the $y$ term to the left side: $\frac{d y}{d x} + \frac{1}{x} y = e^{x - 2 x - 2}$
Compare with standard form: $\frac{d y}{d x} + P (x) y = Q (x)$
Identify $P (x) = \frac{1}{x}$
Identify $Q (x) = e^{x - 2 x - 2}$

Integrating Factor

Formula for IF: $e^{\int P (x) d x}$
Substitute $P (x)$ : $e^{\int \frac{1}{x} d x}$
Rewrite as power: $\int x^{- \frac{1}{2}} d x$
Integrate: $2 x^{\frac{1}{2}} = 2 x$
Result: $I F = e^{2 x}$

General Solution Setup

Formula: $y \cdot I F = \int (Q (x) \cdot I F) d x + C$
Substitute IF and $Q (x)$ : $y \cdot e^{2 x} = \int (e^{x - 2 x - 2} \cdot e^{2 x}) d x + C$

Simplifying the Integrand

Look at the product inside the integral: $e^{x - 2 x - 2} \cdot e^{2 x}$
Add the exponents: $x - 2 x - 2 + 2 x$
The $- 2 x$ and $+ 2 x$ cancel out.
Simplified integrand: $e^{x - 2}$

Integrating the Right Side

We now need to integrate: $\int e^{x - 2} d x$
The integral of $e^{x - 2}$ is simply $e^{x - 2}$
Update the equation: $y \cdot e^{2 x} = e^{x - 2} + C$

Finding the Constant $C$

Use the given condition: $y (0) = 0$
Substitute $x = 0$ and $y = 0$ into the equation.
Left side: $0 \cdot e^{20} = 0$
Right side: $e^{0 - 2} + C = e^{- 2} + C$
Equation: $0 = e^{- 2} + C$

Specific Solution

Solve for $C$ : $C = - e^{- 2}$
Substitute $C$ back into the general solution.
$y \cdot e^{2 x} = e^{x - 2} - e^{- 2}$
Isolate $y$ : $y = \frac{e ^{x - 2} - e ^{- 2}}{e ^{2 x}}$

Evaluating $y (1)$

We need to find the value of $y$ when $x = 1$ .
Substitute $x = 1$ : $y (1) = \frac{e ^{1 - 2} - e ^{- 2}}{e ^{21}}$
Simplify the exponents: $y (1) = \frac{e ^{- 1} - e ^{- 2}}{e ^{2}}$

Final Answer

Rewrite with positive exponents: $y (1) = \frac{\frac{1}{e} - \frac{1}{e ^{2}}}{e ^{2}}$
Find a common denominator for the numerator: $\frac{e - 1}{e ^{2}}$
Divide by $e^{2}$ : $y (1) = \frac{e - 1}{e ^{4}}$
This matches Option 3.

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The problem presents a nested function structure that initially appears complex. We are given

f (x) = x - 1

and

g (x) = e^{x}

To simplify the term

g (f (f (x)))

, we first evaluate the inner composition:

f (f (x)) = f (x - 1) = (x - 1) - 1 = x - 2

Applying the function

g

to this result, we obtain:

g (f (f (x))) = e^{x - 2}

The Master Equation

We now substitute this into the given differential equation:

\frac{d y}{d x} = e^{- 2 x} \cdot e^{x - 2} - \frac{y}{x}

Rearranging the terms to isolate the linear differential form

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

, we get:

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

Here, we identify the components:

P (x) = \frac{1}{x}, Q (x) = e^{x - 2 x - 2}

The Integrating Factor

The Integrating Factor (

I F

) is calculated as follows:

I F = e^{\int P (x) d x} = e^{\int x^{- 1/2} d x} = e^{2 x}

Multiplying the entire differential equation by the

I F

, the left side simplifies to the derivative of the product

y \cdot e^{2 x}

\frac{d}{d x} (y \cdot e^{2 x}) = e^{x - 2 x - 2} \cdot e^{2 x}

Simplifying the right side by adding the exponents:

\frac{d}{d x} (y \cdot e^{2 x}) = e^{x - 2}

Final Calculation

Integrating both sides with respect to

x

y \cdot e^{2 x} = \int e^{x - 2} d x = e^{x - 2} + C

Using the initial condition

y (0) = 0

0 \cdot e^{0} = e^{0 - 2} + C ⟹ 0 = e^{- 2} + C ⟹ C = - e^{- 2}

The general solution is therefore:

y \cdot e^{2 x} = e^{x - 2} - e^{- 2}

To find

y (1)

, we substitute

x = 1

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

Dividing by

e^{2}

, we reach the final result:

y (1) = \frac{e ^{- 1} - e ^{- 2}}{e ^{2}} = \frac{\frac{1}{e} - \frac{1}{e ^{2}}}{e ^{2}} = \frac{e - 1}{e ^{4}}

The final answer is $\frac{e - 1}{e ^{4}}$ .

Common Traps & FAQs

Wait, the integral of

x^{- 1/2}

2 x^{1/2}

, but where did the constant of integration go in the Integrating Factor?

That is a fantastic observation! In the formula for the Integrating Factor,

I F = e^{\int P (x) d x}

, we typically omit the constant of integration. Why? Because the Integrating Factor is a multiplier we use to make the left side of our differential equation a perfect derivative. Adding a constant there would just scale the entire equation by a constant factor, which would cancel out anyway when we solve for

y

. So, we keep it simple and set

C = 0

for the IF calculation.

The exponent

e^{x - 2 x - 2}

looks impossible to integrate. Did I do something wrong?

Take a deep breath! You haven't done anything wrong. In JEE Advanced problems, whenever you see a terrifying exponential term, look for a way to simplify it using the Integrating Factor. When you multiply

Q (x)

by the IF, the exponents add up. Notice how the

- 2 x

from the

Q (x)

term and the

+ 2 x

from the IF perfectly annihilate each other? It is designed to be elegant. Trust the process!

Is the constant

C

always zero?

No, never assume

C = 0

! Always use the initial condition provided in the problem. Here, we were given

y (0) = 0

. We substitute

x = 0

and

y = 0

into our general solution to find the specific value of

C

. In this problem, it turned out to be

- e^{- 2}

. Always respect the initial condition; it is the key that unlocks the specific solution from the general family of curves.

Animated Solution for Mathematics - Differential Equations: Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x}), y (0) = 0$ , then $y (1)$ is :-

Select Answer:

Visualized Solution

Problem Setup

Evaluating $f (f (x))$

Evaluating $g (f (f (x)))$

Updating the Differential Equation

Standard Linear Form

Integrating Factor

General Solution Setup

Simplifying the Integrand

Integrating the Right Side

Finding the Constant $C$

Specific Solution

Evaluating $y (1)$

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

The Integrating Factor

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x−1 and g(x)=ex for x∈R. If dxdy​=(e−2x​g(f(f(x)))−x​y​),y(0)=0, then y(1) is :-

Select Answer:

Visualized Solution

Problem Setup

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(f(x)))

Updating the Differential Equation

Standard Linear Form

Integrating Factor

General Solution Setup

Simplifying the Integrand

Integrating the Right Side

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

Specific Solution

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

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

The Integrating Factor

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Differential Equations: Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x}), y (0) = 0$ , then $y (1)$ is :-

Evaluating $f (f (x))$

Evaluating $g (f (f (x)))$

Finding the Constant $C$

Evaluating $y (1)$

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The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

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

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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 :

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