5

2: Convert the following higher order IVPs into first order systems dy dy mg dt? sin(y) = Asin(Qt) with y(0) = 8.Y(0) = @p: dt b) d7 49-3&9+c*62-- e"'...

Question

2: Convert the following higher order IVPs into first order systems dy dy mg dt? sin(y) = Asin(Qt) with y(0) = 8.Y(0) = @p: dt b) d7 49-3&9+c*62-- e"' sin(3t) , with y(l) = 1,y() = 2,y"(1) = 0. dt'

2: Convert the following higher order IVPs into first order systems dy dy mg dt? sin(y) = Asin(Qt) with y(0) = 8.Y(0) = @p: dt b) d7 49-3&9+c*62-- e"' sin(3t) , with y(l) = 1,y() = 2,y"(1) = 0. dt'



Answers

Which of the following differential equations are first-order?
(a) $y^{\prime}=x^{2} \quad$ (b) $y^{\prime \prime}=y^{2}$
(c) $\left(y^{\prime}\right)^{3}+y y^{\prime}=\sin x \quad$ (d) $x^{2} y^{\prime}-e^{x} y=\sin y$
(e) $y^{\prime \prime}+3 y^{\prime}=\frac{y}{x} \quad$ (f) $y y^{\prime}+x+y=0$

Here. We want to see that for each of the following second order differential equations find at least one particular solution. You will need to call on past experience with functions you have differentiated for a significantly greater challenge. Find the general solutions. So let's just start with part A that says the second derivative of Y equals X and to find a solution to find a why that works. All we have to do is integrate twice. So if we integrate, once we get do I d. X or the first derivative equals X squared over to right now, we're just gonna find one particular solution. Now, if we integrate again to find why we get X cubed over six. So there we go. We have a general solution. We're sorry, a specific solution. But if we want to find a general solution, I'm gonna do it right here. What we have to do is we have to take into consideration what it says that the general solution will involve two unknown constants. So let's integrate. We get the first derivative equals X squared over two, and then we cannot forget a plus c. Let's call it C one Now if we integrate again, we get y equals X cubed over six plus C one x plus another constant C two. So that is a general solution. Now let's do a couple of the other parts. Part B wants us to find the second derivative equals negative X to find a solution. All we have to do is we already different. We already took the integral of the positive of this. We did X. So if we want to do negative X, all we really have to do is just add a negative. So we can really just go ahead and say that the general solution is why equals negative X cubed over six? Um and then we can also just include R plus C one X plus C. Two, because these will never change when you're integrating twice because these are just the result of constants. So that is our general solution for the second derivative equals negative X. Now let's get another page for part C. We have the second derivative equals negative sine of X. So now let's take the integral. If we take the integral, we get the second or the first derivative of Y equals, um co sign of X. I'm gonna say plus c one and if we integrate again, we get y equals the interval of coastline of exits Just sign of X y equals sine of x plus C one x plus our next constant C two. So that is our general solution for part c carte de were given the second derivative of y equals. Why now? This may seem really challenging at first, but let's think about what we've learned about, um during isn't in a rolls for a special case of a function where when you take the derivative, it stays the same, so that should remind us of e to the X. So when we take the derivative index, we have e to the X we integrated index, we have e to the X, so this is a really unique function when we talk about in calculus. So what we can say is, if we know that the second derivative equals d function itself. If we want to rewrite the function, why we can rewrite it as y equals, it could be any constant. We could have a constant c one e to the X because if we have five either the X The derivative is still 58 x we integrate. We saw five e to the X, so the constant doesn't change anything. And then we could have another. Um, we have to have another part of this. We could have C two e to the negative X. So why do we have this? Because we have to take into account that if we take the derivative twice of e to the negative s x were given back to the negative X like, let's write e to the negative x the first derivatives negative e to the negative X. But the second derivative, which is what we're dealing with here, is eating the negative X. So the second derivative equals the first derivative. So we also have to have a term that has even the negative X as the second negative equals, the function holds true. So this is the most difficult one, I think. But that is a general solution right there. And then let's do part E that says, uh, the second derivative equals the negative of the function. So we're gonna have to think hard about this one to this one's pretty difficult, but what is a function that we know when we take a dirty dog? Um, twice. It's going to be the negation of, um, really itself. So I think for me what comes to mind is our trig functions like Sign and Co sign. So let's think about if we have why just This is what I call sign. We have y equals sign and take the derivative we get the first derivative equals co sign, and the second derivative equals negative of science. So that should tell us something. And then let's try. Let's try and said Co sign We have y equals co sign. We have the second derivative of the first relative equals negative sign. And then we have the second derivative equals negative co sign so that we see for sign and co sign. Their second narratives are the allegations of themselves. So there we go. We figured it out. We can say that why equals because of constant again, just like when we dealt with either the X doesn't matter. Why equals c one sine of X plus c two co sign of X because both sign and co sign when you take the derivative and then take the second derivative, the second derivative equals and negation of the function itself. So that is our general solution, and that is that's all the parts of this question and yeah.

Right we want to solve by separation of variables. The equation Y prime is equal to P T X over Y squared this equation Y prime equals E. To the X over Y squared involving a derivative of Y is called the differential equation. And the separation of variables method is one such method to solve these equations. There's two steps we have to follow to solve for this. First we're gonna isolate the variables Y and X on each side of the equation. So put wind left X on the right and then we're going to integrate both sides and solve for Y. We want our final solution to be in terms of why to the first power or one Y. So let's go ahead and isolate our variables since Y prime is dy dx we can write why square do I is equal to dx dx Next we're gonna integrate both sides. So performance integration give us on the right why cucumber three equals E X plus C one. C one is our constant of integration, solving in terms of why give the solution, why is equal to the cube root of three E v x plus three C. One. We can also call three C one simply see and we have our solution here.

Uh huh. We want to solve by separation of variables. The equation why prime is equal to X times E. To the X. All over two. Y. When equations such as this one involving both Y and its derivatives. Known as a differential equation. The separation of variable method is one such method for solving these equations, there are two necessary steps. We have to complete separation of variables here to solve. First, we're gonna isolate our terms or variables on either side of the equation. So put terms of why? On the left, in terms of X on the right, then we'll integrate both sides. Once we integrate both sides, we solve for Y as a function of X. To finish the problem. So first and foremost it's isolate our variables since Y. Prime can be written as dy dx we can isolate our variables. S. Two. Why do I is equal to X? The X. D. X. Next we integrate both sides. So integration gives us on the right. Why squared equals B. to the x x minus one plus. See the constant immigration integration we want to solve in terms of why? So we take the square root to obtain solution. Why is equal to the square root of BD x times x minus one plus six.

Yeah. So for a question area have divide over the X dressy calls to the integrate off 12 X plus four. So d y over the X just equals 26 Squire plus four x with a constant. So integrate both sides. We'll have y equals two to act to the power of three plus two x choir. Let's see one x You forgot the X here, Reese constantly to and for B We'll have integrate off the term Amanda life that that's just equals to the y over the X So we can just write you I over the x here integrate offi to paragraph X sign next the X so D y over the X just equals two. With our constant and now less integrate, both sides will have white just equals two c one x with C two. Yeah, and foresee We'll have the y over the X equals to integrate off extra power of three actual power off minus three. So do you buy over The X just equals two x two powerful over full minus 1/2 extra power to plus the constant. And we integrate both sides. Okay, we can see that it's the integrate off the y. And this is the integrate. Yeah, See, Wan T X. So why Jesse kills Thio acts the power off 5/20 Last one hour two x and constancy.


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