If $f(x)$ is a differentiable function such that $f(x)+f'(x)\leq 1$ for all $x$ and $f(0)=0$ .then largest possible value of $f(1)$ is
solution i tried $e^xf(x)+f'(x)e^x\leq e^x$ $\displaystyle \frac{d}{dx}(f(x)e^x)\leq e^x$
integrating both side $e^xf(x)\leq e^x+c$ $f(x)\leq 1+ce^{-x}$ how to find largest $f(1)$
On
We can treat as equality as a limit of inequality in view if the monotonous nature of the function $y=f(x)$
$$ y+ y^{'}=1 $$
integrates to
$$ 1-y= c e^{-x} $$ Boundary condition $ x=0,y=0$ gives $c=1$
$$ y= 1- e^{-x} $$ @ $x=1$ the function value is
$$ 1-\frac{1}{e} $$
So the function should be at most this value. That is $< = $ the above.
Instead of integrating indefinitely, take a definite integral of both sides.
We have
$$ e^xf(x) + f'(x)e^x = \frac{d}{dx}(f(x)e^x) \leq e^x$$
Integrating both sides from $0$ to $1$ gives
$$ f(1)e^1 - f(0)e^0 \leq e^1 - e^0$$ $$ f(1)e - f(0) = f(1)e \leq e - 1$$ $$f(1) \leq \frac{e-1}e$$