Find the solution subject to the initial conditions.
dP/dt = -2P P(0)=1
dP/-2P = dt
then do i take the antiderivative? what would i do next??
Yes.
ln P = -2 t + C
P = e^(-2t + C)
= C' e^-2t where C' is a different arbitrary constant
C' = 1 since P(0) = 1
Therefore P = e^-2t
What if P= Ce^at + D
dP/dt= aCe^at and if D is zero..
dP/dt= ac e^at=aP
now if a is -2
dp/dt=-2P Hmmm.
P(O)=1=Ce^a0=C so C=1
Now for your question...
dP/P=-2 dt
lnP=-2t
P=e^-2t
thank you!!
To solve this problem, you need to separate the variables and integrate both sides of the equation.
Starting from:
dP/dt = -2P
You can rearrange the equation as follows:
dP/P = -2dt
Now, you can integrate both sides with respect to their respective variables:
∫(dP/P) = -2∫dt
The integral of 1/x with respect to x is ln|x| + C, where C is the constant of integration. The integral of a constant with respect to t is simply the constant multiplied by t.
Therefore:
ln|P| = -2t + C
Now, to find the constant of integration (C), you can use the initial condition P(0) = 1. Substituting this into the equation:
ln|1| = -2(0) + C
0 = 0 + C
C = 0
So, the equation becomes:
ln|P| = -2t
Now, to solve for P, you can raise both sides of the equation as exponentials of base e:
e^(ln|P|) = e^(-2t)
P = e^(-2t)
Finally, you have obtained the solution to the differential equation subject to the given initial condition:
P(t) = e^(-2t)