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Verify that $$y=\sin 3 t+2 \cos 3 t$$ is a solution to the initial value problem

$$2 y^{\prime \prime}+18 y=0 ; \quad y(0)=2, \quad y^{\prime}(0)=3$$

Find the maximum of $$|y(t)| \text { for }-\infty<t<\infty$$.

$|y|_{\max }=\sqrt{5}$

05:13

A M.

Calculus 2 / BC

Chapter 4

Linear Second-Order Equations

Section 1

Introduction: The Mass-Spring Oscillator

Differential Equations

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Oregon State University

Harvey Mudd College

Baylor University

Lectures

01:11

In mathematics, integratio…

06:55

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Verifying solutions of ini…

04:32

Sketch the solution to the…

08:02

Solve the given initial-va…

00:59

Find a rectangular equatio…

00:51

03:47

Eliminate the parameter t.…

02:18

01:03

01:32

03:23

$y^{\prime \prime}+y=3 \si…

01:22

02:39

01:50

Solve the initial-value pr…

03:38

10:16

02:58

Use the Laplace transform …

03:37

Verify that $ y = -t \cos …

03:05

$$y^{\prime \prime}-2 y^{\…

So what we see is, um, that y of zero is equal to two, which is going to equal sign of three times zero plus to co sign of three T or three of zero, actually, three times zero. So that's gonna equal, as we know. Sign up 00 and co. Sign of zeros one. So that's two times one, which equals two. So, in fact, we do see that y zero equals two. So that initial condition satisfied? Then when we take my prime, we get three co sign three t plus six, actually minus six, sign three t. Then we see that y prime of zero is equal to, um, when we plug a zero in here, we get three times one minus six times zero. So we get three, which is what we predicted for initial conditions. So we verified that, um, and then what we want to do one more time is take the double derivative. So y double prime is going to end up giving us a three, actually, nine Kerr sane. Ah, let's restart this negative nine sign three T minus 18, co sign three t and then we're gonna add that now so to y prime is going to give us negative 18 sign three T minus 36 co sign three team and then we want to add this to positive 18 y Well, we see that 18 y is going to give us 18 sign three t minus are actually plus 36 co sign three t. So you see that when we add these together, these are going to cancel. These are going to cancel, leaving us with zero, which is exactly once again what is a solution to the initial value problem, Then, in finding the maximum for y of t for negative infinity to infinity To find that maximum, we let the derivative function equals zero. And since we know that its initial value is three, um, when we let this equal zero, we can end up seeing that as a result, when we use our trick properties, we get that Ry max is equal to the square root of five

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