Use existence and uniqueness theorem to prove that the solution of an initial-value problem for the equation $x'=\dfrac{x}{1+t^2}$ with x(0) 0 can never become negative.
Hint: First find a constant solution of the differential equation for some constant C.
Please, help

Question

Use existence and uniqueness theorem to prove that the solution of an initial-value problem for the equation $x'=\dfrac{x}{1+t^2}$ with x(0) >0 can never become negative.
Hint: First find a constant solution of the differential equation for some constant C.
Please, help

loser66 · Answer

f(t, x) is continuous every where and $\dfrac{\partial}{\partial x}(f(t,x))=\dfrac{1}{1+t^2}$ is continuous every where also. Hence the theorem guarantee that the solution exists and unique.

loser66 · Answer

However, I don't get the part "never become negative" . I don't know what I am supposed to do

amilapsn · Answer

you can easily solve it and get $\Large x=Ce^{tan^{-1}t}$.  x(0)>0  implies C>0 implies x>0 implies x can never be negative... Am I right?

amilapsn · Answer

Without solving it you can get the result too.