Let $f$ be a smoth strictly positive function on the real line. Define $X(t,a)$ to be the solution of the differential equation $$x'(t)=f(s,x(s)),\\x(a)=0.$$ The question is: How to prove that the solution $a\rightarrow x(t,a)$ is decreasing for all $t$ greater then $a$?. I was thinking of writing the solution in the integral form $$x(t,a)=\int_a^tf(s,x(s,a))ds,$$ since the solution is given implicitly, we can not compute the derevative with respetc to $a$.
Any suggestion?. Thank you in advance.
You need to show that for $a<b<t$ you get $x(t,a)>x(t,b)$.
Now because of $f>0$ you get $x(b,a)>0=x(b,b)$. By the uniqueness theorem, solutions can not cross, so that $x(t,a)>x(t,b)$ for all $t$ where both solutions are defined.