Question

Suppose a function E:R → R is defined as the solution of the ODE E'(x) = -TE(), E(0) = 1. We will assume that this equation has a solution, and that Ex) #0 for all x E R. For this problem, you are to answer all the equations without solving the differential equationforget that you might be able to do this! (a) Prove that E(x) > 0 for all x (recall: we assume E(x) + 0). (b) Use the mean value theorem to prove that E() is strictly increasing for < < 0 and strictly decreasing for x > 0. (c) Prove that E is bounded above. (d) Prove that limz-40 E(x) = 0. (e) Use the identity criterion to show that the function h(x) = ln(E(+ 3x2 is identically 0. Given this, what is E(x)?

Answer 1

solution:

E:IR$\rightarrow$ IR E'(x) = - xE(x) ,E(0)

a)

E'(x) $\Rightarrow$ xE(x) = 0 $\Rightarrow$ x=0 ($\because$E(x) $\neq$ 0

E(0) = 1 > .sp if E($\alpha$)=0 , which is contraduction thus E(x)>0  $\forall$x e IR.

b)

E',(x)= - xE(x)

Since E(x).o   $\forall$ x

so, if x <0 $\Rightarrow$ E'(x)>0 $\Rightarrow$ E is strically increasing

c)

since E is monotonically decreasing for E is

E (0) = s $\Rightarrow$ E(x) $\leqslant$ 1 $\forall$ x>0

As E is monotonically increasing for x<0,

$\Rightarrow$ E(x) is bounded above

d)

As E decrease for x>0

so asx$\rightarrow$$\infty$ E(x) $\rightarrow$ 0

i. e

lim0o E() = 0. lim

please give me thub up