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What is a solution to the differential equation #dP-6Pdt=0# with P=5 when t=0?

Answer 1

Cameron Alexander

#P = 5 e^{6t}#

#dP-6Pdt=0#

so separate it!!

#dP = 6Pdt#

#(dP)/P = 6dt#

#int (dP)/P = 6 int dt#

#ln P = 6t + alpha#

#P = e^{6t + alpha} = e^{6t} e^{ alpha} = beta e^{6t}#

NB: we have just replaced the constant #e^{ alpha}# by #beta#, a simpler form of constant, so #beta = e^{ alpha}#

with final form #P = beta e^{6t}#, we can plug in the initial value so

#5 = beta e^{6 times0} = beta#

so we have

#P = 5 e^{6t}#

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Answer 2

Emily Ammerman

To find a solution to the given differential equation ( \frac{dP}{dt} - 6P = 0 ) with the initial condition ( P(0) = 5 ), we can separate variables and integrate.

Separating variables:

[ \frac{dP}{P} = 6dt ]

Integrating both sides:

[ \int \frac{1}{P} dP = \int 6 dt ]

[ \ln|P| = 6t + C ]

Where ( C ) is the constant of integration.

Now, applying the initial condition ( P(0) = 5 ):

[ \ln|5| = 6(0) + C ]

[ \ln|5| = C ]

So, the particular solution is:

[ \ln|P| = 6t + \ln|5| ]

[ \ln|P| = \ln|5| + 6t ]

[ P = e^{\ln|5| + 6t} ]

[ P = 5e^{6t} ]

Thus, the solution to the differential equation with the initial condition ( P(0) = 5 ) is ( P = 5e^{6t} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.