great videos , you have talent in teaching and delivering knowledge...i salute you ....in the first example i think the solution should be : P=Ae^(Bt)such that AxB=K....
Except it isn't. Here's why: Start with your equation: P = A*e^(B*t) Take the derivative relative to t: dP/dt = A*B*e^(B*t) Our goal is for dP/dt to equal k*P. So does it? A*B*e^(B*t) =?= k*A*e^(B*t) Cancel the A, cancel the e^(B*t). We are left with: B =?= k This could only be a solution to the original equation, in the special case that B=k, which would happen when A=1. Eddie's solution is correct. P=A*e^(k*t) is the solution to dP/dt = k*P. Take its derivative and show that it satisfies the diffEQ: dP/dt = A*k*e^(k*t) A*k*e^(k*t) =?= k* (A*e^(k*t)) Cancel the A, cancel the k, cancel e^(k*t), and we're left with, 1=?=1, which confirms this equation is the correct solution.
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Eddie bro you're a lifesaver
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great videos , you have talent in teaching and delivering knowledge...i salute you ....in the first example i think the solution should be : P=Ae^(Bt)such that AxB=K....
Except it isn't.
Here's why:
Start with your equation:
P = A*e^(B*t)
Take the derivative relative to t:
dP/dt = A*B*e^(B*t)
Our goal is for dP/dt to equal k*P.
So does it?
A*B*e^(B*t) =?= k*A*e^(B*t)
Cancel the A, cancel the e^(B*t). We are left with:
B =?= k
This could only be a solution to the original equation, in the special case that B=k, which would happen when A=1.
Eddie's solution is correct. P=A*e^(k*t) is the solution to dP/dt = k*P.
Take its derivative and show that it satisfies the diffEQ:
dP/dt = A*k*e^(k*t)
A*k*e^(k*t) =?= k* (A*e^(k*t))
Cancel the A, cancel the k, cancel e^(k*t), and we're left with, 1=?=1, which confirms this equation is the correct solution.
@@carultch YOU GENIUS
great video sir, ik im kinda 10 years late but the glare was a bit bad XD
thankyou