I missed the day of class where we went over solving 2nd order diff-eqs.
y''(t) - 100 y(t) = 0
subject to y(0)=100 and y'(0)=3000
Thank you!
integrate both sides
$\displaystyle y'-50y^2=a$................(i)
again integrate
$\displaystyle y-\frac{50}{3}y^3=ay+b$.......................(ii)
now put t=0 in eq (i)&(ii)
3000-50(10000)=a
100-(50/3)*1000000=100a+b
solve it for a and b and pun in eq (ii) to get the req ans
$\displaystyle y''(t) - 100y(t) = 0$
This is second order linear constant coefficient homogeneous.
So the characteristic equation is
$\displaystyle m^2 - 100 = 0$
$\displaystyle m^2 = 100$
$\displaystyle m = \pm 10$.
So the solution is
$\displaystyle y(t) = c_1e^{10t} + c_2e^{-10t}$.
You have two initial conditions: $\displaystyle y(0) = 100$ and $\displaystyle y'(0) = 3000$.
From the first, we can see
$\displaystyle 100 = c_1e^{10(0)} + c_2e^{-10(0)}$
$\displaystyle 100 = c_1 + c_2$.
From the second...
$\displaystyle y(t) = c_1e^{10t} + c_2e^{-10t}$
$\displaystyle y'(t) = 10c_1e^{10t} - 10c_2e^{-10t}$
$\displaystyle 3000 = 10c_1e^{10(0)} - 10c_2e^{-10(0)}$
$\displaystyle 3000 = 10c_1 - 10c_2$
$\displaystyle 300 = c_1 - c_2$.
So now you have two equations in two unknowns:
$\displaystyle 100 = c_1 + c_2$
$\displaystyle 300 = c_1 - c_2$.
Adding them together, we find
$\displaystyle 100 + 300 = (c_1 + c_2) + (c_1 - c_2)$
$\displaystyle 400 = 2c_1$
$\displaystyle c_1 = 200$.
Back substituting:
$\displaystyle 100 = c_1 + c_2$
$\displaystyle 100 = 200 + c_2$
$\displaystyle c_2 = -100$.
So finally, the solution is
$\displaystyle y(t) = 200e^{10t} - 100e^{-10t}$.
Sorry, but this won't work.
$\displaystyle y''(t) - 100y(t) = 0$
$\displaystyle y''(t) = 100y(t)$.
If you were going to integrate both sides, you would have to integrate with respect to $\displaystyle t$, not $\displaystyle y$. And since you don't know anything about $\displaystyle y$ yet, you can't possibly integrate it with respect to $\displaystyle t$.