If a, b, and c are all positive constants and y(x) is a solution of the differential equation a

If a, b, and c are all positive constants and y(x) is a...

Key Concepts

Homogeneous Linear Differential Equations

Homogeneous linear differential equations feature terms that are solely dependent on the function and its derivatives, with no additional non-homogeneous terms present. They allow solutions to be constructed as linear combinations of functions determined by the equation's structure, typically through the method of characteristic equations when the coefficients are constant.

Characteristic Equation

The characteristic equation arises by substituting an exponential trial solution into the differential equation. It is a polynomial equation whose roots determine the form of the general solution. In cases where the coefficients are positive, the roots' real parts are crucial because they govern the decay or growth of the solution.

Stability and Exponential Decay

Stability in differential equations refers to the tendency of the solutions to approach a steady state or zero as time increases. When the characteristic roots have negative real parts, the associated exponential functions decay to zero as x approaches infinity, ensuring that the overall solution approaches zero. This is the key reasoning behind why y(x) approaches 0 in the given equation.

Transcript

00:01 We ask to show that this second order differential equation, a, y, double prime, plus b, y, prime, plus c, y, equals zero.

00:08 Always has a solution that goes to zero as the independent variable, say, x, goes to infinity if a, b, and c are positive.

00:22 So then, i don't know if there's any clever way of doing that by taking limits of the differential equation, but we can just look at the solutions.

00:30 So we know the solutions of the form a, e, the lambda, x.

00:37 And plugging that in, we get the equation.

00:43 A, lambda squared plus b, lambda plus c, equals zero.

00:48 Solving for lambda, we get that equals negative b plus or minus square a b squared of b squared minus 4a c all over 2a, just the quadratic formula.

00:57 And so what you can see here is that we get solutions that look like lambda equals, equals c1 plus or minus c2.

01:08 I'm just gonna call it that.

01:10 So then we have e to the lambda and that we can write as e to the c1, e to the plus or minus c2.

01:18 So we have two solutions.

01:20 And so here this is our general form here.

01:23 So we can see we can pull these out, right? and so we can see c1 is always negative, right? c1 is always negative.

01:34 So that means that this is always, exponentially decaying.

01:37 But nc2 is always positive.

01:43 So that's just this part here, and that's when taking the positive square root here, is either always positive or imaginary.

01:55 If it's imaginary, then these are bounded by one.

02:01 So we have exponentially decaying always.

02:05 If c, if, so these are bound.

02:07 If c1, if c2 is imaginary, then these are, are signs and cosines and these are bounded...

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