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let $f:\mathbb R \to \mathbb R$ be a differentiable function with $f(0) = 1$ and satisfying the equation $$f(x+y) = f(x)f '(y) + f '(x)f(y)\qquad \forall ~x,y \in \mathbb R$$ then find the value of $\quad \ln \bigl (f(4)\bigl) $

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3 Answers 3

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The first step to try is to fix one of the variables. For instance, set $y=0$ to get $$ f(x)=f(x)f'(0)+f'(x) $$ which is a simple differential equation. Then insert the solution into the full equation to fix the occurring constants.

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Obviously

$$ f(0) = 2f'(0) $$

which implies $f'(0) = \tfrac{1}{2}.$ Together with $$ f(x) = \tfrac{1}{2}f(x) + f'(x) $$

we get a trivial ODE $f'(x) = \tfrac{1}{2} f(x)$ with $f(0) = 1$.

Now you can easily compute the Value of $\ln(f(4))$.

Please recognise, that your post doesn't even contain a question. Think about this fact, provided that this section is called questions...

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The function $~f~$ satisfies $$f(x+y) = f(x)f '(y) + f '(x)f(y)\qquad \forall ~x,y \in \mathbb R$$

Putting $~x=y=0~$ in the above relation we have $$f(0) = f(0)f '(0) + f '(0)f(0)\implies f'(0)=\frac{1}{2}$$

So the required function satisfies $f(0) = 1\quad \text{and}\quad f'(0)=\frac{1}{2}$.

Let $$~f(x)=e^{\frac{x}{2}}$$

Clearly $~f~$ is differentiable in $~\mathbb{R}~$ and $$~f'(x)=\frac{1}{2}~e^{\frac{x}{2}}.$$

Now $$f(0) =e^{0}= 1\quad \text{and}\quad f'(0)=\frac{1}{2}~e^{0}=\frac{1}{2}$$

also $$~f(x)f '(y) + f '(x)f(y)=e^{\frac{x}{2}}\cdot \frac{1}{2}~e^{\frac{y}{2}}+\frac{1}{2}~e^{\frac{x}{2}}\cdot e^{\frac{y}{2}}=e^{\frac{x+y}{2}}=f(x+y).$$

So our consideration is correct.

Now $$\quad \ln \bigl (f(4)\bigl)= \ln \bigl (e^{\frac{4}{2}}\bigl)=2.$$

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