Show that $g'(x)+g(x)-2e^x=0$

Question

Given a function $g$ which has derivative $g'$ for all x $\in {R}$ and satisfying $g'(0)=2$ and $g(x+y)=e^yg(x)+e^xg(y)$ for all $x,y\in {R}$

Show that $g'(x)+g(x)-2e^x=0$

$\dfrac{g(x+y)}{e^{x+y}}=\dfrac{g(x)}{e^{x}}+\dfrac{g(y)}{e^{y}}$

Putting $x=0$,

$g'(y)=2e^y+g(x)$

Also putting $y=0$,we get,

$0=e^xg(0)\implies g(0)=0$

I had a strong hunch that $g(x)=e^x-e^{-x}$ but it does not satisfy $g(x+y)=e^yg(x)+e^xg(y)$

Please help.

Answer 1

$$g(x+y) = e^y g(x) + e^x g(y)$$
Differentiating w.r.t. y
$$g'(x+y) = e^y g(x) + e^x g'(y)$$
Now put $y=0$
$$g'(x) = e^0 g(x) + e^x g'(0)\implies g'(x) - g(x) - 2e^x = 0$$

Answer 2

Let $h(x)=\frac{g(x)}{e^x}$. Then, $h'(x)=\frac{g'(x)e^x-g(x)e^x}{e^{2x}}$.

Note that you wrote $h(x+y)=h(x)+h(y)$.

Then, $h'(x)=\lim_{\epsilon\to 0} \frac{h(x+\epsilon)-h(x)}{\epsilon}=\lim_{\epsilon\to 0}\frac{h(\epsilon)}{\epsilon}=h'(0)=2$ whatever $x\in\mathbb{R}$ since you noted that $h(0)=g(0)=0$.

then $h'(x)=\frac{g'(x)e^x-g(x)e^x}{e^{2x}}$ becomes $2e^x+g(x)=g'(x)$. Any chance that sign was the other way?

Answer 3

As noted in question and one other answer, with $h(x)=e^{-x}g(x)$ you get $h$ differentiable at $x=0$ and $h(x+y)=h(x)+h(y)$. This is one of Cauchy's fundamental functional equations with solution $h(x)=ax$. Now insert to $g(x)=axe^x$ and compare the derivatives $g'(x)=a(1+x)e^x$ at $x=0$, $2=g'(0)=a$, to get the unique solution.

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Setting $x=0$ in the original equation gives $$ g(y)=e^yg(0)+g(y)\implies g(0)=0 $$ Computing the $x$-derivative and then setting $x=0$ results in $$ g'(x+y)=e^yg'(x)+e^xg(y)\implies g'(y)=e^yg'(0)+g(y)\implies g'(y)-g(y)-2e^y=0 $$ which has one sign different than the ODE that you obtained.

Answer 4

As pointed out in other answers/comments, if $g'(x)\color{red}{-}g(x)-2e^x=0, g'(0)=2$, then: $$g(x)=2xe^x \Rightarrow g(x+y)=e^yg(x)+e^xg(y).$$