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Show that if $a, b,$ and $c$ are positive constants, then all solutions of the homogeneous differential equation$$a y^{\prime \prime}+b y^{\prime}+c y=0$$approach zero as $x \rightarrow \infty$.

Calculus 3

Chapter 17

Second-Order Differential Equations

Section 1

Second-Order Linear Equations

Johns Hopkins University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

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Given positive constants, a b and c, we are asked to consider this differential equation and we will show case by case that the solutions to that differential equation all approach 0 as x approaches infinity. For starters, the auxiliary or characteristic equation is a r squared plus b r plus c equals 0, and this equation, as solutions are 1 or 2, and throughout this problem i am going to refer to this differential equation. So i'm gonna label it as star and for our first case we have b squared minus 4, a c equals 0 point. Then, by theorem 4 from section 17.1, the solution to star is y equals c 1 e to the negative b over 2, a times x, plus c 2 times x times e to the negative b over 2 a x, and here it should be clear that both Of these terms, they approach 0 as x, approaches infinity. So for this 1 in particular, the first term, we have e to a negative, constant times x, so this approaches 0 as x, approaches infinity and then the same is true for this as well. That is this term. This approaches 0 as x, approaches infinity and to show that you can use lupito's rule right. This is x over e to the b over 2 a times x, so the exponential winds over the linear, polynomial x now for case 2. In this case, we're considering what happens if b, squared minus 4 a c is less than 0 point, then, by theorem 5 from section 17.1, the solution of the star is e to the alpha times x times this expression containing cosines and signs where alpha is negative B over 2 a and then thetis, this expression so also notice in this case, we'll have complex roots, that is r 1 and r 2. There will be imaginary. Numbers are complex numbers, okay, and even in this case, we still have that the limit of y equals 0 as x approaches infinity right. This function, approaches 0 since cosine and sine are bounded right. They'Re bounded by their absolute value, is bounded by 1 and then we have either the alpha x, which alpha this number is less than 0. So since cosine and sin are bounded and we have e to some negative constant times x, that's going to approach 0 as x, approaches, infinity and, lastly, is case 3. So, in this case, we're considering what happens when b, squared minus 4 a c is greater strictly greater than 0, then by theorem 3 from section 17.1, the solution to star is y equals c 1, either the r 1 x plus c 2 e to the r 2 x and just as a reminder right stars, this differential equation, which we keep, which we're solving and then our 1 and r 2 are these values here. So let me actually write them down again now. What we want to show is that both of these terms, approach 0 as x, approaches infinity n now for 1 of them, it's very clear- and that is this term right here- this s, 2 e to the r 2 times x right. It'S clear that this approaches. 0 as x, approaches infinity because the number or 2 this is a negative number, so b, is negative and then minus the square root of b squared minus 4 a c. This is also a negative number all right. So r 2 itself is a negative number and it turns out that r 1 is also a negative number as well recall that a and c are greater than 0 point. So then, if i take the product of 2 positive numbers that will still be greater than 0, i can multiply it by 4 and then negative 4, a c. This is less than 0 point and what i can do is add, b squared to both sides and that doesn't change the sign of the inequality. Then i can take the square root of both sidesand. On the rate we just have, b since b b is greater than 0. We can take the square root, nothing weird happens. Okay, then i can move b to the other sideand. This is less than 0 and i can divide both sides by 2, a of course, 0 divided by 2, a is still just 0, but by dividing by 2, a this term on the left is exactly equal to r onepoiright, so in particular, r 1 is less Than 0- and this implies that the limit as x approaches infinity of c 1 e to the r 1 x, plus c, 2 e to the r 2 x. This is equal to 0 right and that's because both are 1 and our 2 are less than 0.

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