Question

1. Find the particular solution to the differential equation, given the general solution is

y =√9t + C and the initial condition that the solution curve passes through (2, 18).

PLEASE HELP ASAP

2. Use separation of variables to find the solution to the differential equation dP/dt = 6P if P = 15 when t = 1

Answer 1

1. If you mean

`y = √(9t) + C`

then from the initial condition we find

`18 = √(9\cdot2) + C \implies C = 18 - 3\sqrt2`

If you mean

`y = √(9t + C)`

then

`18 = √(9\cdot2 + C) \implies 18^2 = 18 + C \implies C = 306`

2. Separating variables, we have

`(dP)/(dt) = 6P \implies \frac{dP}P = 6\,dt`

so that upon integrating, we get

`\displaystyle \int \frac{dP}P = 6 \int dt \implies \ln|P| = 6t + C`

Given that
`P=15` when
`t=1`, we find

`\ln|15| = 6\cdot1 + C \implies C = \ln(15) - 6`

and so

`\ln|P| = 6t + \ln(15) - 6 \\\\ \implies e^(\ln|P|) = e^(6t + \ln(15) - 6) \\\\ \implies \boxed{P = 15e^(6t-6)}`