Standard reduction of order procedure: suppose there is a second solution of the form

, which has derivatives



Substitute these terms into the ODE:



and replacing

, we have an ODE linear in

:

Divide both sides by

, giving

and noting that the left hand side is a derivative of a product, namely
![\dfrac{\mathrm d}{\mathrm dx}[wx]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Bwx%5D%3D0)
we can then integrate both sides to obtain


Solve for

:


Now

where the second term is already accounted for by

, which means

, and the above is the general solution for the ODE.
Factors of 14: 7 and 2
7 + 2 = 9
The correct equation is 8.7 + b = 54.6
because it is given that measure of side a is 8.7cm, in all other equations the values of a is different.
from the first equation we can find the value of b, that is
b = 54.6 - 8.7 = 45.9
so, value of b is 45.9cm
It would be 5³ because 5×5×5= 125
he is going to need 37.5 ounces I guess