Try This One on for Size

For the second-order differential equation we use the auxiliary equation $m^2+m=0$ and we will solve for $m$ . The two roots of the equation will be the exponents of the solution $y=C_1 e^{m_1}+C_2 e^{m_2}$ .

From here, we can solve the quadratic auxiliary equation finding the two solutions

$m_1, \: m_2 = \frac{-1 \pm \sqrt{1^2-4 \times 1 \times 0}}{2 \times 1} \: \Rightarrow \: m_1=0, \: \: \textrm{and} \: \: m_2=-1$

So now we have the solution $y=C_1+C_2 e^{-x}$ and now we need to plug in those boundary conditions. Well first, we can find that $y'=-C_2 e^{-x}$ and we can solve for $C_2$ . Plugging in our second initial condition $y'( \ln 2 )=1$ we see that $C_2=-2$ .

Now we can plug in our first initial condition, much like we did the second and we find that $C_1=C_2=-2$ . We finally have our solution to the ODE.

$f(x)=-2 \left ( e^{-x} +1 \right )$

Now let's plug in $\ln 4$

$f( \ln 4 ) =-\frac{5}{2}$ and therefore $a+b =5+2=\boxed{7}$