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Find the solution to the differential equation that satisfies the initial conditions y(0)=50
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where is the differential equation?
dy/dt=ky
k is a constant
hmmm separable equation?
1/y dy = k dt... ln y = k t y = e^kt ?
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how do you include the initial condition?
it's been a year since my calculus ab course and now i'm in physics c :/
Y(0) = 50 soo the ?? 50 = e^k(0) ???
that doesn't look right ... that's 1 = 50 nooooo
ln y = k t
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e^lny = e^kt djfslakjdaljfd
would it be y=50e^k*t then 100=50e^k*5 so k=0.1386 y=50e^90.1386)8t??
that 8 should be *
y+50e^(0.1386)*t
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