Why did we set $x=0$ and $y=0$ at the start? Is that always allowed?

Yes, because the equation $f(x+y) = f(x)f'(y) + f'(x)f(y)$ holds for all $x, y \in R$. It is like having a master key; $x=0$ and $y=0$ are the simplest values that unlock the hidden constants within the function.

Can we just assume $f(x) = e^{kx}$?

While that is a clever intuition for exponential growth, in JEE Advanced, you must prove it. Deriving the differential equation is the rigorous path that ensures you have not missed any other possible solutions or constraints.

Why did we drop the absolute value in $\ln|y|$?

Great observation! Since $f(0)=1$ and the function is differentiable (and thus continuous), $f(x)$ must remain positive for all $x$. Therefore, $|f(x)|$ is simply $f(x)$.

JEE Advanced 2018

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a differentiable function with $f (0) = 1$ and satisfying the equation $f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)$ for all $x, y \in R$ . Then, the value of $lo g_{e} (f (4))$ is ________.

Enter Numerical Value:

Visualized Solution

The Functional Equation

Given Functional Equation:
$f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)$
Initial Condition:
$f (0) = 1$
Goal: Find $lo g_{e} (f (4))$

Finding $f^{'} (0)$ Setup

Substitute $x = 0$ and $y = 0$ into the equation:
$f (0 + 0) = f (0) f^{'} (0) + f^{'} (0) f (0)$

Calculating $f^{'} (0)$

$f (0) = 2 f (0) f^{'} (0)$
Substitute the known value $f (0) = 1$ :
$1 = 2 (1) f^{'} (0)$
$f^{'} (0) = \frac{1}{2}$

Generating a Differential Equation

Substitute $y = 0$ into the original equation:
$f (x + 0) = f (x) f^{'} (0) + f^{'} (x) f (0)$

Substituting Known Values

We know $f (0) = 1$ and $f^{'} (0) = \frac{1}{2}$ .
Substitute these into the equation:
$f (x) = f (x) \cdot \frac{1}{2} + f^{'} (x) \cdot 1$

Isolating the Derivative

Subtract $\frac{1}{2} f (x)$ from both sides:
$f^{'} (x) = f (x) - \frac{1}{2} f (x)$
$f^{'} (x) = \frac{1}{2} f (x)$

Separation of Variables

Rewrite $f^{'} (x)$ as $\frac{d y}{d x}$ and $f (x)$ as $y$ :
$\frac{d y}{d x} = \frac{1}{2} y$
Separate the variables:
$\frac{d y}{y} = \frac{1}{2} d x$

Integrating Both Sides

$\int \frac{1}{y} d y = \int \frac{1}{2} d x$
$ln ∣ y ∣ = \frac{x}{2} + C$
Since $f (0) = 1 > 0$ , we can drop the absolute value:
$ln (f (x)) = \frac{x}{2} + C$

Evaluating the Constant $C$

Use the initial condition $f (0) = 1$ .
Substitute $x = 0$ :
$ln (f (0)) = \frac{0}{2} + C$
$ln (1) = 0 + C$
$0 = C ⟹ C = 0$

The Explicit Function

Substitute $C = 0$ back into the equation:
$ln (f (x)) = \frac{x}{2}$
Taking the exponential of both sides:
$f (x) = e^{\frac{x}{2}}$

Evaluating $lo g_{e} (f (4))$

We need to find $lo g_{e} (f (4))$ , which is $ln (f (4))$ .
Using our earlier equation: $ln (f (x)) = \frac{x}{2}$
Substitute $x = 4$ :
$ln (f (4)) = \frac{4}{2} = 2$
Final Answer: $2$

00:00 / 00:00

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

We are given a differentiable function

f : R \to R

satisfying the functional equation:

f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)

We are also provided with the initial condition

f (0) = 1

. Our objective is to determine the value of

lo g_{e} (f (4))

The Zero Strategy

In functional equations, the origin is often the key to unlocking the structure. Let us set

x = 0

and

y = 0

in the given equation:

f (0 + 0) = f (0) f^{'} (0) + f^{'} (0) f (0)

This simplifies to the expression:

f (0) = 2 f (0) f^{'} (0)

Given

f (0) = 1

, we substitute this value into the equation:

1 = 2 (1) f^{'} (0) ⟹ f^{'} (0) = \frac{1}{2}

The Bridge to Calculus

To determine the explicit form of

f (x)

, we transform the functional equation into a differential equation. We keep

x

as a variable and set

y = 0

f (x + 0) = f (x) f^{'} (0) + f^{'} (x) f (0)

Substituting

f (0) = 1

and

f^{'} (0) = \frac{1}{2}

, we obtain:

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

Rearranging the terms leads to a first-order linear differential equation:

f^{'} (x) = f (x) - \frac{1}{2} f (x) = \frac{1}{2} f (x)

The Integration Journey

We now solve the differential equation

\frac{d y}{d x} = \frac{1}{2} y

using the method of separation of variables:

\frac{d y}{y} = \frac{1}{2} d x

Integrating both sides yields:

ln ∣ f (x) ∣ = \frac{x}{2} + C

Applying the initial condition

f (0) = 1

, we find

ln (1) = 0 + C

, which implies

C = 0

. Thus, the function is defined by:

ln (f (x)) = \frac{x}{2}

Final Calculation

The problem asks for the value of

lo g_{e} (f (4))

, which is equivalent to

ln (f (4))

. Using our derived relation

ln (f (x)) = \frac{x}{2}

, we substitute

x = 4

ln (f (4)) = \frac{4}{2} = 2

The final answer is 2.

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a differentiable function with $f (0) = 1$ and satisfying the equation $f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)$ for all $x, y \in R$ . Then, the value of $lo g_{e} (f (4))$ is ________.

Enter Numerical Value:

Visualized Solution

The Functional Equation

Finding $f^{'} (0)$ Setup

Calculating $f^{'} (0)$

Generating a Differential Equation

Substituting Known Values

Isolating the Derivative

Separation of Variables

Integrating Both Sides

Evaluating the Constant $C$

The Explicit Function

Evaluating $lo g_{e} (f (4))$

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Zero Strategy

The Bridge to Calculus

The Integration Journey

Final Calculation

Similar Questions

Let $f (x)$ be a positive function such that the area bounded by $y = f (x), y = 0$ from $x = 0$ to $x = a > 0$ is $e^{- a} + 4 a^{2} + a - 1$ . Then the differential equation, whose general solution is $y = c_{1} f (x) + c_{2}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is

Let $y = f (x)$ be a curve passing through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves.

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is:

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The differential equation representing the family of curves $y^{2} = 2 c (x + c)$ , where $c$ is a positive parameter, is of

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The differential equation for the family of circle $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

Enter Numerical Value:

Visualized Solution

The Functional Equation

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

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

Generating a Differential Equation

Substituting Known Values

Isolating the Derivative

Separation of Variables

Integrating Both Sides

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

The Explicit Function

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } loge​(f(4))

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Zero Strategy

The Bridge to Calculus

The Integration Journey

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation of the family of curves, x2=4b(y+b), b∈R, is:

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation representing the family of curves y2=2c(x+c​), where c is a positive parameter, is of

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation whose solution is Ax2+By2=1 where A and B are arbitrary constants is of

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Finding $f^{'} (0)$ Setup

Calculating $f^{'} (0)$

Evaluating the Constant $C$

Evaluating $lo g_{e} (f (4))$

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is:

The differential equation satisfied by the system of parabolas $y^{2} = 4 a (x + a)$ is:

The differential equation which represents the family of curves $y = c_{1} e^{c_{2} x}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is

The differential equation of the family of circles passing through the points $(0, 2)$ and $(0, - 2)$ is

The differential equation representing the family of curves $y^{2} = 2 c (x + c)$ , where $c$ is a positive parameter, is of

The differential equation whose solution is $A x^{2} + B y^{2} = 1$ where $A$ and $B$ are arbitrary constants is of

The differential equation for the family of circle $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is