2/3 = x/1...2 parts water to 3 parts juice = x parts water to 1 part juice
cross multiply
(3)(x) = (2)(1)
3x = 2
x = 2/3 parts water are needed for 1 part juice
next 3 steps:
ABC<u>DC</u>BA
ABCD<u>ED</u>CBA
ABCDE<u>FE</u>DCBA
Patern: add 2 letters
1. 1
2. 3
3. 5
4. 7
5. 9
6. 11
7. 13
8. 15
9. 17
10. 19
The 10th step has 19 letters
Substitute

, so that

![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes


which has the characteristic equation

with roots at

. This means the characteristic solution for

is

and in terms of

, this is

From the given initial conditions, we find


so the particular solution to the IVP is
5k3 - 3k + 7 - (<span>-2k3 + k2 - 9)
= </span> 5k3 - 3k + 7 + 2k3 - k2 + 9
= 7k3 - k2 - 3k + 16