9 a solve the initial value problem in terms of alpha y 3y...

9. (a) Solve the initial value problem (in terms of \alpha ): y′′ – 3y′ – 18y = 0, y(0) = –12, y′(0) = \alpha . (b) If lim y(t) = 0, what is the value of \alpha ? t -> \infty

Transcript

00:01 Consider the following differential equation.

00:04 Y ' -2y ' -3y is equal to 0.

00:08 In our first question, question a, we want to find the general solution.

00:12 So let's pose our general solution as y equal to the exponential of lambda t.

00:20 Differentially y two times and inserting the results into our differential equation, we obtain the following characteristic equation.

00:27 Equation.

00:28 Lambda squared minus 2 lambda minus 3 is equal to 3.

00:33 Now i want to solve for lambda using a quadratic formula.

00:43 We find that this factors quite nicely as lambda plus 1 times lambda minus 3, plus 3, excuse me.

00:56 Oh no, i had it, minus 3, which means that lambda is either equal to minus 1 or 3.

01:04 So now our general solution y of t will correspond to the linear combination of what we have posed.

01:12 So y will be equal to c1 times the exponential of minus t plus c2 times the exponential of 3t.

01:29 Now we are given an initial condition.

01:32 At t equal to 0, y is equal to alpha and y prime is equal to 1.

01:37 We want to find the values of alphas for which we have a finite solution.

01:47 So essentially, we have initial value problems now.

01:51 An initial value problem.

01:53 So let's evaluate y at 0.

01:58 And we obtain c1 plus c2.

02:03 And our initial condition tells us that y at 0 is equal to alpha.

02:09 Let's compute y prime...

Question

9. (a) Solve the initial value problem (in terms of \alpha ): y′′ – 3y′ – 18y = 0, y(0) = –12, y′(0) = \alpha . (b) If lim y(t) = 0, what is the value of \alpha ? t -> \infty

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