A. Solve the following initial value problem: (t^2

A. Solve the following initial value problem: (t^2 - 10t + 16) dy/dt = y with y(5) = 1. (Find y as a function of t.) y = . B. On what interval is the solution valid? Answer: It is valid for < t < . C. Find the limit of the solution as t approaches the left end of the interval. (Your answer should be a number or the word "infinite".) Answer: . D. Similar to C, but for the right end. Answer: .

Transcript

00:01 Okay, we want to solve this differential equation with this not exactly initial condition, but it's a condition on the solution okay, and so we can separate variables and then actually that factors into y minus 8 and y minus 2 okay now the next step is to put it as a to do partial fractions decomposition of this, but there's a trick we got to realize first and that is when we do this integration we're going to get logarithms and so when we're done when we get to that point we want to make sure then we put in our initial condition it's not going to give us the log of a negative number and right now if i was to put in y equals 5 the right -hand side would be negative so what i want to do is i want to change i want to rewrite the equation so it looks like this so i get minus dy over y is dt over 8 minus y times y minus 2 that way the right -hand side if i put in y equals 5 is positive okay, and then i'll do my partial fractions so i get a over 8 minus y plus b over y minus 2 okay, and so and that's a t excuse me again like that, okay? and and that's times dt so then i'm gonna get um i multiply that out i get a times t minus 2 plus b times 8 minus t that has to equal 1 based on what's on the left -hand side there over my denominator okay, so that tells me that a minus b equals 0 and 8 b minus 2 a equals 1 so we got a equals b and so we get a equals b equals 1 6th so then we get dy over y or the minus sign is 1 6th of a 1 1 over 8 minus t plus 1 over t minus 2 okay, now if i integrate that and there's a dt so now i integrate this i get minus the log of y on the left -hand side and on the right -hand side i get 1 6th times the log of t minus 2 minus the log of 8 minus t okay, the reason for that minus sign is that i got a minus t in the denominator here so the differential of this is minus dt so i can work with this little so i get 1 over y because that's minus log y i take the exponential of the whole thing and i'm going to get a plus log of a constant over here so i get c times the logarithm of t minus 2 over 8 minus t raised to the 1 6th power and there's no logarithm i'm taking the exponential of both sides so i get that there is my expression for y as a function of t okay, i'm not done yet though if i put in t equals 5 he has to give me y equals 1 so i get 1 equals c times i get 3 over 3 raised to the 1 6th power that tells me c equals 1 so that y i'm gonna flip that over i get 8 minus t over t minus 2 to the 1 6th power there's my answer okay, and then we want to look at what the limits are.

06:15 So right now based on the logarithms that i had i expect that my t i'm gonna have 2 less than t less than 8 but so if i put in t equals 2 then i'm gonna get infinity here...

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