# Thread: Solve this 2nd order IVP!!

1. ## Solve this 2nd order IVP!!

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!

2. Originally Posted by tanelly
y''(t) - 100 y(t) = 0
subject to y(0)=100 and y'(0)=3000
integrate both sides
$y'-50y^2=a$................(i)
again integrate
$y-\frac{50}{3}y^3=ay+b$.......................(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

3. Originally Posted by tanelly
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!
$y''(t) - 100y(t) = 0$

This is second order linear constant coefficient homogeneous.
So the characteristic equation is

$m^2 - 100 = 0$

$m^2 = 100$

$m = \pm 10$.

So the solution is

$y(t) = c_1e^{10t} + c_2e^{-10t}$.

You have two initial conditions: $y(0) = 100$ and $y'(0) = 3000$.

From the first, we can see

$100 = c_1e^{10(0)} + c_2e^{-10(0)}$

$100 = c_1 + c_2$.

From the second...

$y(t) = c_1e^{10t} + c_2e^{-10t}$

$y'(t) = 10c_1e^{10t} - 10c_2e^{-10t}$

$3000 = 10c_1e^{10(0)} - 10c_2e^{-10(0)}$

$3000 = 10c_1 - 10c_2$

$300 = c_1 - c_2$.

So now you have two equations in two unknowns:

$100 = c_1 + c_2$

$300 = c_1 - c_2$.

$100 + 300 = (c_1 + c_2) + (c_1 - c_2)$

$400 = 2c_1$

$c_1 = 200$.

Back substituting:

$100 = c_1 + c_2$

$100 = 200 + c_2$

$c_2 = -100$.

So finally, the solution is

$y(t) = 200e^{10t} - 100e^{-10t}$.

4. Originally Posted by slovakiamaths
integrate both sides
$y'-50y^2=a$................(i)
again integrate
$y-\frac{50}{3}y^3=ay+b$.......................(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
Sorry, but this won't work.

$y''(t) - 100y(t) = 0$

$y''(t) = 100y(t)$.

If you were going to integrate both sides, you would have to integrate with respect to $t$, not $y$. And since you don't know anything about $y$ yet, you can't possibly integrate it with respect to $t$.