1. Interpret the given conditions

We are given a thrice differentiable odd function $f:\mathbb{R}\to\mathbb{R}$ such that

$f''(x)=f(x), \qquad f(0)=0, \qquad f'(0)=3.$

(Also $f'(x)\ge 0$ is given, which will be automatically satisfied by the solution we obtain.)

We need to find:

$9f(\ln 3).$

2. Solve the differential equation

Given

$f''(x)=f(x),$

the auxiliary equation is

$m^2=1 \implies m=\pm 1.$

So the general solution is

$f(x)=Ae^x+Be^{-x}.$

3. Use the condition that $f$ is odd

Since $f$ is odd,

$f(-x)=-f(x).$

Now,

$f(-x)=Ae^{-x}+Be^x.$

For oddness,

$Ae^{-x}+Be^x=-(Ae^x+Be^{-x}).$

Comparing coefficients of $e^x$ and $e^{-x}$, we get

$B=-A.$

Hence

$f(x)=A(e^x-e^{-x})=2A\sinh x.$

4. Use the initial condition $f'(0)=3$

Differentiate:

$f'(x)=A(e^x+e^{-x})=2A\cosh x.$

At $x=0$,

$f'(0)=2A=3 \implies A=\frac{3}{2}.$

Therefore,

$f(x)=\frac{3}{2}(e^x-e^{-x}).$

Also,

$f(0)=\frac{3}{2}(1-1)=0,$

so the condition $f(0)=0$ is satisfied.

5. Compute $f(\ln 3)$

Substitute $x=\ln 3$:

$f(\ln 3)=\frac{3}{2}\left(e^{\ln 3}-e^{-\ln 3}\right).$

Now,

$e^{\ln 3}=3, \qquad e^{-\ln 3}=\frac{1}{3}.$

So,