Given a differentiable function $f$ with certain conditions, prove that $f(x) = e^x + 1$ for every $x \in \mathbb{R}$.

Question

Given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is differentiable and given that $f(0)=2$ and $(e^x + 1)f'(x) = e^x f(x)$ for every $x \in \mathbb{R}$, prove that $f(x) = e^x + 1$.

Answer 1

Hint

$${d\over dx} \ln f(x)={f'(x)\over f(x)}$$

Answer 2

That's a pretty straightforward "differential equation". From $(e^x+ 1)f'(x)= (e^x+ 1)\frac{df}{dx}= e^xf(x)$ we have $\frac{df}{f}= \frac{e^x}{e^x+ 1} dx$.

Now integrate both sides. On the left we have $ln(f)+ C_1$. To integrate on the right, let $u= e^x+ 1$. Then $du= e^xdx$ so the right side becomes $\frac{du}{u}$. Integrating we have $ln(u)+ C_2= ln(e^x+ 1)+ C_2$.

Then $ln(f)+ C_1= ln(e^x+ 1)+ C_2$ so $ln(f)= ln(e^x+ 1)+ (C_2- C_1)$ or, letting $C'= C_2- C_1$, $ln(f)= ln(e^x+ 1)+ C'$.

Take the exponential of both sides: $f(x)= e^{ln(e^x+ 1)+ C'}= e^{ln(e^x+ 1)}e^{C'}= e^{C'}(e^x+ 1)$.

Finally, letting $C= e^{C'}$. $f(x)= C(e^x+ 1)$.

Since we are told that f(0)= 2, setting x= 0 in that equation, $f(0)= C(e^0+ 1)= C(1+ 1)= 2C= 2$ so $C= 1$.

We have $f(x)= e^x+ 1$.

Check: $f(0)= e^0+ 1= 1+ 1= 2$. $f'(x)= e^x$ so $(e^x+ 1)f'(x)= (e^x+ 1)e^x= e^{2x}+ e^x$. And $e^xf(x)= e^x(e^x+ 1)= e^{2x}+ e^x$. Yes, $(e^x+ 1)f'(x)= e^xf(x)$.

(Strictly speaking, there should be a whole lot of "absolute values" in those logarithms, such as $ln|e^x+ 1|$, and $e^{C'}$ would be positive for any C'. But allowing C to be positive or negative then fixes the absolute value problem.)