Differential Equations Question involving $f(x+y)$ [closed]

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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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.

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 $$\ln(f(4))$$.

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