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Numerade Educator



Problem 6 Medium Difficulty

$$\begin{array}{ll}{3 x^{n}+5 x-2 y=0 ;} & {x(0)=-1, \quad x^{\prime}(0)=0} \\ {4 y^{\prime \prime}+2 y-6 x=0 ;} & {y(0)=1, \quad y^{\prime}(0)=2}\end{array}$$


$x_{2}^{\prime}(t)=\frac{-5 x_{1}+2 x_{3}}{3}$
$x_{4}^{\prime}=\frac{6 x_{1}-2 x_{3}}{4}$
$x_{1}(0)=-1, x_{2}(0)=0$
$x_{3}(0)=1, x_{4}(0)=2$


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Video Transcript

Okay, so we have the following system of different equations. So we have three. Why? Double prime plus find X and then minus two. Why is equal to zero? Sorry. Three x three extra crime. Those five X minus two by equal zero. And then we also have four. Why? Double prime Plus two. I minus six x is equal to zero. So this is to second order differential equations. And we would like to convert this system into a system of four first order differential questions. And the way we do this is is similar to what we've done before. We just start assigning names to all of our variables and their derivatives. But first, let's write down. Actually, the initial conditions we have is, well, we have x zero is equal to you. Negative one. Yeah. Ex prime of zero equals zero. Why? Zero is equal to one. And why prime of zero is equal to two. Okay, so here's what I'm gonna do. I'm gonna set x one to be ex X to to be ex prime x three to be why and then x four to be why prayer? So this first equation becomes three x two. Prime because X double crime plus five x one minus two x three equals zero. Then for the second equation, we have four x four Prime cause Why double Prime plus two x three minus six x one equals zero and then our final two differential equations actually relate Ex one next to in X three texts for and that's that x one prime is equal to X to and x three. Prime is equal to X for So there are four first order differential equations with our initial conditions. That x 10 His negative one x 20 zero x three of zero is one and x 40 equals two. So we've converted these two second order differential equations and 24 1st order, different equations.