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Problem 3

$$\begin{array}{l}{y^{(4)}(t)-y^{(3)}(t)+7 y(t)=\…

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Problem 2

$$\begin{array}{l}{y^{\prime \prime}(t)=\cos (t-y)+y^{2}(t)} \\ {y(0)=1, \quad y^{\prime}(0)=0}\end{array}
$$

Answer

$x_{1}^{\prime}(t)=x_{2}(t)$
$x_{2}^{\prime}(t)=\cos \left(t-x_{1}(t)\right)+x_{1}(t)^{2}$
$x_{1}(0)=1$
$x_{2}(0)=0$


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Video Transcript

All right. So we have a differential equation. Second order. Why? Double crime is equal. Thio co sign of T minus y and then plus y squared. And we have some initial conditions in the initial conditions. Are that why? So here, wise, of course. A function of tea. And if we evaluate why at syrup we get one. And if we evaluate why prime and zero, we get zero. And what we would like to do is convert this differential equation into a system. Uh, first order equations, OK, And the way you do that is really simple. So you just say x once and my first variable is just going to be Why in my second variable is going to be why prime? And we only have to do this up into until one less than the derivative that gives the order of the differential equation. So in this case, since it's the second order, we only need two variables. And now notice that we have one equation that's really easy to see, So we have that X one prime. That's just why prime is equal to X to. So this is one of our for shorter equations And then our next equation is gonna come from our original equation that why double prime is X to crime and then we have co sign it's okay to have a team, but why? We want to write his ex one. And then again we have x one square. So what we just did is we converted this first or the second order differential equation into a system of two first word or equations and now our conditions on ex one next to are the X one, which is why zero is equal to one and then x, too, Which is why prime of zero equal zero. So here is what's called the standard form of this system of differential equations.

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