Okay, so let's start by finding the solution to this differential equation. So I will do that by first finding the auxiliary equation socially, B R squared plus four R plus seven. And set that equal to zero. So now, um, let's see if we can factor this. So, for example, seven, um, has on Lee the roots or only the factors one and seven, which does not add up to four. So therefore, we're gonna need to use quadratic formula. We're gonna get our signal to negative B and then plus or minus square root of B squared, which is 16 minus four times a, which is one time see, which is seven eso we get, uh, here. Uh, okay. All over two A, which is just too so out front. Here we have. Um that's gonna become just negative too, then plus or minus. Now we have 16 minus 14 here, so that's gonna be square root of to. So we get squared of 60 minus 14 Which is this positive too? So we'll have square root of 2/2. Ah, here. Oh, wait. Sorry, Actually, um, four times. Sorry. Four times seven is 28. So we actually have 16 minus 28. So 16 minus 28. That's going to be equal to, um, square to 12. So we're gonna have square root of 12 or negative 12/2. So we're gonna actually have so I square root of 12/2 square to 12 can become to square root of three over two. So we'll just get I square root of three. So that's gonna become I square root of three like so now, since we have imaginary roots, imaginary roots means that we're going to be under dense here, So under damped. Okay, So since we have an under dam system with Graf oven under dam system, it's going to look like an oscillation that goes to zero or with AMFA's Who that goes to zero. So we started, say to We're going ah, up first. So we started to go up and then let me do that in different color, actually, again started to then go up, and then we're going to all sleep till we get, um, our oscillations get smaller and smaller here. Now, for our actual solution, we're going to have first that we're gonna have the General solution. Wire T is gonna be able to see one e to the negative two teeth and then co sign, and then we're gonna have square root of three t Let me cried as t square root of three and then plus C to eat of the late of two t times Sign Ah, tee Times Square with a three. So then now Ah, actually, I'll rewrite it. A square to three t square root of three and then times t square root of three. I'm Steve. Okay, so now we need to use our, um, initial conditions. So if we plug in zero first into this sign of 00 to 0, and then we have C one e to the zero co sign a zero both of these Air one. So we gets get C one here. So why of zero? I think the C one is able to to so we can just plug into here, So that's just gonna be a hate to now we can. Or now we need to find the derivative. Why? Prime of tea. Okay, so we're going t to use product rule a couple times here. Okay. Right. Okay. So when we take this derivative first let's do Ah, first times derivative second. So that's gonna be equal to negative Two heats of the negative to t uh, and then times also square root of three. So we'll have to square to three and then sign square root of three t and then also will have minus four e to the negative to t and then co sign square to three t Then for the second half, we're gonna have a plus plus so first each of the NATO to t times driven of second. So then we're gonna need to add a square to three and then, um, cosigned square root of three t and then now ah, minus two c to easily native to t And then, um, leave that a sign square root of three t like So now the sign terms are just gonna go away when we plug in zero. So why prime of zero is equal to again. So these go to zero. This goes to zero and then Oh, sorry, this goes to zero and this goes to zero. So we're left with. This is negative four because this goes to one. This goes to one negative four and C two square root of three. So we have negative four plus seat to square root of three is equal to six. So we're gonna have C two is gonna be with you 10 divided by square root of three or 10 square root of 3/3. Like so. So our final solution. I'm gonna move this over here. Our final solution is gonna look like why of tea is equal to to eat a plate of two teeth, cosigned square root of three t and then plus 10 square root of 3/3 E to the negative to t time sign square root of three t like so. So that's our final, um, motion.