Originally Posted by
Chris L T521 Hmm...this is what I think you should do.
I suggest you solve the differential equation $\displaystyle \frac{\,dy}{\,dx}=0.05y$ with the initial condition $\displaystyle y(0)=1000$
By separation of variables, this leads to $\displaystyle \frac{\,dy}{y}=0.05\,dx$
Integrate both sides, and we see that $\displaystyle \int\frac{\,dy}{y}=0.05\int\,dx\implies \ln\left|y\right|=0.05x+C\implies\left|y\right|=e^ {0.05x+C}$ $\displaystyle \implies \left|y\right|=e^Ce^{0.05x}\implies y=Ke^{0.05x};~K=\pm e^{C}$
Now applying the initial condition $\displaystyle y(0)=1000$, you get $\displaystyle 1000=Ke^{0.05\cdot0}\implies K=1000$
Thus, your function is $\displaystyle y\!\left(x\right)=1000e^{0.05x}$
Now go ahead and find $\displaystyle \frac{\,d^2y}{\,dx^2}$
Does this make sense? Can you take it from here?