JEE Advanced 2023

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $f : [1, \infty) \to R$ be a differentiable function such that $f (1) = \frac{1}{3}$ and $3 \int_{1}^{x} f (t) d t = x f (x) - \frac{x ^{3}}{3}, x \in [1, \infty)$ . Let $e$ denote the base of the natural logarithm. Then the value of $f (e)$ is

Q: Wait, why do we differentiate? Can't we just integrate the right side directly?

The right side contains $f(x)$, which is an unknown function. We cannot integrate $f(x)$ without knowing its form. Differentiation is the only way to transform the integral into a standard differential equation that we can solve. It is a powerful tool to 'free' the function from the integral.

Q: How do I know it's a Linear Differential Equation?

Look at the structure: we have a term with $f'(x)$ and a term with $f(x)$. Since there are no higher powers like $(f(x))^2$ or $(f'(x))^2$, it is linear. The standard form $\frac{dy}{dx} + Py = Q$ is your roadmap for solving these types of equations.

Select Answer:

Visualized Solution

The Integral Equation

Given Equation:
$3 \int_{1}^{x} f (t) d t = x f (x) - \frac{x ^{3}}{3}$
Goal: Find the value of $f (e)$

Applying Leibniz Rule

To remove the integral, differentiate both sides with respect to $x$ .
$\frac{d}{d x} (3 \int_{1}^{x} f (t) d t) = \frac{d}{d x} (x f (x) - \frac{x ^{3}}{3})$

Executing Differentiation

Left Hand Side (Leibniz Rule): $3 f (x)$
Right Hand Side (Product Rule): $1 \cdot f (x) + x \cdot f^{'} (x) - \frac{3 x ^{2}}{3}$
Equating both sides: $3 f (x) = f (x) + x f^{'} (x) - x^{2}$

Rearranging the Equation

Subtract $f (x)$ from both sides:
$2 f (x) = x f^{'} (x) - x^{2}$
Rearranging terms:
$x f^{'} (x) - 2 f (x) = x^{2}$

Standardizing to LDE

Divide the entire equation by $x$ (since $x \geq 1$ ):
$\frac{df}{d x} - \frac{2}{x} f (x) = x$

Identifying LDE Components

Standard Linear Differential Equation (LDE) form:
$\frac{d y}{d x} + P y = Q$
Comparing with our equation:
$P = - \frac{2}{x}$
$Q = x$

Integrating Factor Setup

Formula for Integrating Factor (IF):
$I F = e^{\int P d x}$
Substitute $P$ :
$I F = e^{\int - \frac{2}{x} d x}$

Calculating the Integrating Factor

Evaluate the integral: $\int - \frac{2}{x} d x = - 2 ln x$
Substitute back: $I F = e^{- 2 l n x}$
Using log property $a ln b = ln (b^{a})$ :
$I F = e^{l n (x^{- 2})} = x^{- 2} = \frac{1}{x ^{2}}$

General Solution Setup

Formula for LDE solution:
$f (x) \cdot I F = \int (Q \cdot I F) d x + C$
Substitute $I F = \frac{1}{x ^{2}}$ and $Q = x$ :
$f (x) \cdot \frac{1}{x ^{2}} = \int (x \cdot \frac{1}{x ^{2}}) d x + C$

Integrating the RHS

Simplify the integrand: $x \cdot \frac{1}{x ^{2}} = \frac{1}{x}$
Integrate: $\int \frac{1}{x} d x = ln x$
Equation becomes: $\frac{f ( x )}{x ^{2}} = ln x + C$
Multiply by $x^{2}$ : $f (x) = x^{2} ln x + C x^{2}$

Applying Initial Condition

Given condition: $f (1) = \frac{1}{3}$
Substitute $x = 1$ :
$\frac{1}{3} = (1)^{2} ln (1) + C (1)^{2}$
Since $ln (1) = 0$ :
$\frac{1}{3} = 0 + C ⟹ C = \frac{1}{3}$
Particular Solution: $f (x) = x^{2} ln x + \frac{x ^{2}}{3}$

Evaluating f(e)

Substitute $x = e$ :
$f (e) = e^{2} ln (e) + \frac{e ^{2}}{3}$
Since $ln (e) = 1$ :
$f (e) = e^{2} (1) + \frac{e ^{2}}{3}$
$f (e) = \frac{3 e ^{2} + e ^{2}}{3} = \frac{4 e ^{2}}{3}$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

We are given the integral equation:

3 \int_{1}^{x} f (t) d t = x f (x) - \frac{x ^{3}}{3}

The presence of the integral with a variable upper limit

x

suggests the use of the Leibniz Rule. By differentiating both sides with respect to

x

, we transform this integral equation into a differential one.

Transforming into a Differential Equation

Applying the Leibniz Rule to the left side yields

3 f (x)

. Applying the product rule to the right side gives

f (x) + x f^{'} (x) - x^{2}

Equating these results, we obtain:

3 f (x) = f (x) + x f^{'} (x) - x^{2}

This simplifies to the first-order linear differential equation:

x f^{'} (x) - 2 f (x) = x^{2}

Solving the Linear Differential Equation

Dividing the entire equation by

x

(assuming $x eq 0$), we get:

\frac{df}{d x} - \frac{2}{x} f (x) = x

To solve this, we determine the Integrating Factor (IF), defined as

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

. With

P (x) = - \frac{2}{x}

, our

I F

becomes:

I F = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = \frac{1}{x ^{2}}

Multiplying the differential equation by this

I F

allows us to express the left side as the derivative of a product:

\frac{d}{d x} (f (x) \cdot \frac{1}{x ^{2}}) = 1

Final Calculation

Integrating both sides with respect to

x

, we find:

\frac{f ( x )}{x ^{2}} = x + C

Using the initial condition derived from the original integral equation at

x = 1

, we see

3 (0) = 1 \cdot f (1) - \frac{1}{3}

, which implies

f (1) = \frac{1}{3}

. Substituting this into our general solution:

\frac{1/3}{1 ^{2}} = 1 + C \Rightarrow C = - \frac{2}{3}

Thus, the function is defined as:

f (x) = x^{3} - \frac{2 x ^{2}}{3}

Finally, substituting

x = e

, we arrive at the result:

f (e) = e^{3} - \frac{2 e ^{2}}{3}

The final value is $f (e) = e^{3} - \frac{2 e ^{2}}{3}$ .

Common Traps & FAQs

Wait, why do we differentiate? Can't we just integrate the right side directly?

The right side contains

f (x)

, which is an unknown function. We cannot integrate

f (x)

without knowing its form. Differentiation is the only way to transform the integral into a standard differential equation that we can solve. It is a powerful tool to 'free' the function from the integral.

How do I know it's a Linear Differential Equation?

Look at the structure: we have a term with

f^{'} (x)

and a term with

f (x)

. Since there are no higher powers like

(f (x))^{2}

(f^{'} (x))^{2}

, it is linear. The standard form

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

is your roadmap for solving these types of equations.

Why do we need the Integrating Factor?

The Integrating Factor is the 'magic key' that turns the left side of the equation into the derivative of a product, specifically

\frac{d}{d x} [f (x) \cdot I F]

. It allows us to integrate both sides easily and isolate

f (x)

Select Answer:

Visualized Solution

The Integral Equation

Applying Leibniz Rule

Executing Differentiation

Rearranging the Equation

Standardizing to LDE

Identifying LDE Components

Integrating Factor Setup

Calculating the Integrating Factor

General Solution Setup

Integrating the RHS

Applying Initial Condition

Evaluating f(e)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Transforming into a Differential Equation

Solving the Linear Differential Equation

Final Calculation

Similar Questions

Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is

Let $α$ be a non-zero real number. Suppose $f : R \to R$ is a differentiable function such that $f (0) = 2$ and $lim_{x \to - \infty} f (x) = 1$ . If $f^{'} (x) = α f (x) + 3$ , for all $x \in R$ , then $f (- ln_{e} 2)$ is equal to

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

For a differentiable function $f : R \to R$ , suppose $f^{'} (x) = 3 f (x) + α$ , where $α \in R, f (0) = 1$ and $lim_{x \to - \infty} f (x) = 7$ . Then $9 f (- lo g_{e} 3)$ 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:

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 $f$ be a differentiable function in the interval $(0, \infty)$ such that $f (1) = 1$ and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $2 f (2) + 3 f (3)$ is equal to _______

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function, If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of $f (3)$ is :

Select Answer:

Visualized Solution

The Integral Equation

Applying Leibniz Rule

Executing Differentiation

Rearranging the Equation

Standardizing to LDE

Identifying LDE Components

Integrating Factor Setup

Calculating the Integrating Factor

General Solution Setup

Integrating the RHS

Applying Initial Condition

Evaluating f(e)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Transforming into a Differential Equation

Solving the Linear Differential Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→R be a differentiable function. If 6∫1x​f(t)dt=3xf(x)+x3−4 for all x≥1, then the value of f(2)−f(3) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be differentiable on the interval (0,∞) such that f(1)=1, and limt→x​t−xt2f(x)−x2f(t)​=1 for each x>0. Then f(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→x​t−xt2f2(x)−x2f2(t)​=0 If f(x)=1, then x is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a differentiable function f:R→R, suppose f′(x)=3f(x)+α, where α∈R,f(0)=1 and limx→−∞​f(x)=7. Then 9f(−loge​3) 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; } The function y=f(x) is the solution of the differential equation dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1) satisfying f(0)=0. Then ∫−3​/23​/2​f(x)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a differentiable function in the interval (0,∞) such that f(1)=1 and limt→x​t−xt2f(x)−x2f(t)​=1 for each x>0. Then 2f(2)+3f(3) is equal to _______

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function, If 10∫1x​f(t)dt=5xf(x)−x5−9 for all x≥1, then the value of f(3) is :

Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

For a differentiable function $f : R \to R$ , suppose $f^{'} (x) = 3 f (x) + α$ , where $α \in R, f (0) = 1$ and $lim_{x \to - \infty} f (x) = 7$ . Then $9 f (- lo g_{e} 3)$ 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:

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 $f$ be a differentiable function in the interval $(0, \infty)$ such that $f (1) = 1$ and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $2 f (2) + 3 f (3)$ is equal to _______

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function, If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of $f (3)$ is :