solve-the-following-differential-equations-use-your-calculator-to-draw-a-family-of-solutions-are-t-4

Question

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?
$$
x y^{\prime}=2 \frac{\cos x}{x}-3 y
$$

Robert Daugherty · Accepted Answer

Section four, not five. Problem to 36. We&#x27;re dealing with ordinary differential equations, linear first order. So I need to get this in standard forms of the standard form for this New Year equation. It&#x27;s just going to be y prime. Plus, some function of X times y is equal to some other function of X. So getting this in standard form with divide everything by ex some have why prime and then minus And so it&#x27;s gonna be minus three over ex excuse me plus plus three over X. Why is equal to and so divine anger? Actually two over X squared co sign X now in this for money to find my integrating factor. So my integrating factor is going to be e into the anti derivative of three over x dx. So this is E. And so you know that that is three natural log of X that is e natural log of execute that is X cubed, so X cubed is my integrating factor. And so this different your equation can be written as X cubed times why prime is equal to and when you multiply by X cubed. So what you end up with is X cubed times. Juan Prime is equal to two x co sign X And now if I integrate both sides of this equation, I&#x27;m gonna end up with X cubed. Why is equal to when I integrate this? I&#x27;ll have to x sign of X push to co sign of X plus a constant of integration. And then to solve this for why you get wise equal to two over x squared sign of X um, plus two over x cubed co sign of X plus C over X cubed. So that is the family of solutions to this differential equation and narrow Asked to take a look and see what would different conditions be for this eso. If I take a look at this, I could vary that constant of integration. And so you see different changes of the solution based on that. So it looks like there&#x27;s, ah, change. When c is equal to zero, that&#x27;s noticeable. So when I get, uh, C is equal to zero on, it looks like the change happens. Also, another change happens here, where it reflects that C is equal to somewhere around negative to put 12.2. So those were just different conditions. Based on that constant integration that would change boundary conditions for the solution

Robert Daugherty · Answer

Section 45 Problem to 38. We have a first order linear ordinary differential equation and we can solve this by using integrating factors. First thing is to get this in standard form. So standard form is going to be why prime plus some function of X times Why is equal to some other function of X? So if I divide everything about the sign that will give me the standard form here, So I&#x27;m going to have y prime plus co sign of X. We&#x27;re sign of X. Why, If you go to two X over a sign of X now, as with calculus, you got to remember your house. We can&#x27;t remember your trig. So this is why prime this is the co tangent of X times. Why is he going to X over the sign of acts? So are integrating Factor is going to be, uh, meal X. This is equal to E, and that is the anti derivative of co tangent of X, the X. So that is e, uh to the natural log sign of X, which is sign of acts. So this differential equation can be written as sign of X times. Why prime, you&#x27;re multiplying. Everything by sign of X is equal to two X and then the next step, pretty straightforward now is just integrate both sides and you get why times a sign of X is equal to when you integrate. X quits was X squared over two. So that&#x27;s just going to be X squared, plus a constant of integration. And in your final solution, divide by sign of X, you get X squared or a sign of acts I see over sign of X. Or you could choose to write this as y is equal to X squared co seeking acts. It was a constant coast seeking of X, so that would be the family of solutions that solves this different equation. Take a look at that. What does it look like? So you see, different initial conditions would give us different values for that constant C. And then, based on those changes, you see how the solution looks like a, you know, sort of a modified coast seeking. But what changes is are some of the inflection points and intercepts

Robert Daugherty · Answer

Section 45. Problem to 40. We&#x27;re dealing with first order in your ordinary differential equations in this particular section, so everything needs to be in standard forms a standard form for this equation. B y prime some function of X times Why is equal to some other function of X? Get it In that form, I confined. The integrating factor is going to be e to the anti derivative the x, the X. So the places in that former divide everything by X cubed. So you it end up with Why Cube? Why prime? Plus, you&#x27;ve got two X squared about a bags cube that&#x27;s two over X wine is equal to X plus one over excuse, so the integrating factor will be Yeah, the anti derivative of two of Rex T X through This is e to, and that&#x27;s to natural long of X two Antrel Long of X, which is key to the log of X squared, which is simply X squared. So that is my integrating factors. Or multiply this entire equation by the integrating factor that gives me the left side being ah, product room we have why times X squared crying. It&#x27;s equal to So you multiply by X squared. You get X squared over X plus one over X cubed X plus one over X and says, I know I&#x27;m gonna integrate. This was just write it as one plus one over X. So now your solution to equation would be to integrate both sides of this equation. So in doing so on the left side of this equation, you get why except squared that&#x27;s equal to. And when you integrate on the right side of this equation, you would get X after a long of X plus, a constant of integration and in the final step is simply divide by X squared. So X over X squared. That&#x27;s one over X plus long of X over X squared waas. See over expert. So that is the family of solutions that saw this particular different equation. Now take a look at this and see what do different boundary conditions do for this. So the founder could it would change that constant of integration. You look at that constant of integration changes. These would all be the way this solution in the graph would change for the equation

Robert Daugherty · Answer

Section four out five, number 2 35 X y prime And this is actually a plus. So plus 1/2 um, why is it gonna sign of three t&#x27;s? I&#x27;m assuming y is a function of X. So to get this in standard form, this is gonna be wide prime plus 1/2 X. Why is equal to one over acts times a sign of three t The integrating factor is e the integral of 1/2 X the X that is e yeah, to the 1/2. And this is just natural long of X that is e to the natural log square root of X, which is the square root of X. So multiply the entire equation by me an integrating factor, and I&#x27;ll end up with y square root of X prime is equal to the square root of X over X sign of three teeth. Now, if you integrate both sides of this equation, But before we do that, that is what Why Route Ex prime is he Quito. One overt acts sign of three t. So that&#x27;s just X to the minus 1/2 sign of three t. Now we&#x27;re going to integrate both sides of this equation with respect to X, And in doing so, you get why square root of X it&#x27;s equal to So we integrate extra the minus 1/2 You&#x27;re gonna get extra the 1/2 divided by 1/2. So you get two x sign of three T, plus a constant off integration and two X to the sorry, its two extra the 1/2 times the constant of integration. Now you divide everything by X of the 1/2. So you&#x27;re gonna have why is equal to two sign of three t plus c over the square root off X. So that is our family of solutions for this equation. Now let&#x27;s go take a look and see to sign three t plus c over root X Take a look at this and see what different conditions would happen so often Times just go looking to what happens when C is zero. So when C is zero, this is just a, um um see over the square root of X function. And so I don&#x27;t know about this parameter t, so it&#x27;s going to vary. C is going to vary and so you see different variations of this function, um, as the parameter t and the initial boundary condition is what would determine the value of see as that would would change.

Robert Daugherty · Answer

Section four, not five Problem to 34 here were asked to solve an ordinary first order linear differential equation and then look at the graph of the family of solutions. So to put this in standard form, this is going to be y prime plus two. Why is equipped with three e to the t over three and then are integrating factor? It&#x27;s going to be e to the integral two d x. So this is e to the to t. So this problem becomes why e to the to t prime is equal to three e to the t over three either the to t and so this is equal to three and then e toothy 7/3 t. So now the solution for this guy will be to integrate both sides of this equation. And so you end up with why he to the to t is equal to. So when you integrate this, you&#x27;re going to have a three he to the 7/3 t and then you divide by 7/3 which is 3/7 plus a constant of integration. So this gives us why he to the to t is equal to 9/7. e to the 7/3 t plus c And then now to finally solve this for why you&#x27;re gonna have why is equipped on 9/7 e to the 7/3 t minus two t plus C e to the minus two t So 7/3 money, 6/3. So this is why people 9/7 e to the t over three plus ce to the miners to t So that is our family of solutions for this. So if you take a look at the graph here, this is what we see. What does it look like for different values of C? So one of the always good things to do is to look at when CIA zero, when c is zero. He only had the first term. So, hoops, let me get it to be zero when c is zero, you only have that first term and so 9/7 e to the exercise. You see that one exponential every other time you&#x27;ve got the sum of two exponential functions, and so you see the various different behaviors for constants of integration and you see a change there as well. Um, where it starts to go up so These are the family of solutions we could look at that is it animates. But these are the family of solutions for this particular, um, or new to natural question.

Robert Daugherty · Answer

Section 45 problem or to 37 were dealing with ordinary linear first order differential equations here. So everything needs to be put in the standard form with this first order equation. That&#x27;s why prime plus some function of X times Y is equal to some other function of X. Then we confined our integrating factor. So to get this equation in standard form, everything would be divided by X plus one. So you have y prime, and then you get minus three over X plus one. Why is equal to so X square? Plus two experts, one that is X plus one squared, and you may dividing it by X plus one. So this equation is why prime minus three over X plus one. Why is equal to X plus one so to find our integrating factor. So the integrating factor mu of X is going to be E and an anti derivative of negative three over X plus one DX. And so this is E, and this is minus three natural log of X plus one. This is E natural log X plus one to the minus three, which is one over X plus one cubed so that is my integrating factor so I can transform this equation by multiplying by that integrating factor. Because when I do that, the left will set up as the derivative of a product. So you&#x27;ll have. Why, over X Plus one Cubed prime is equal to and on the left side you&#x27;ll have X plus one over X plus one cubed. So that&#x27;s just one over X plus one square. And so now, to solve this, we will integrate both sides of this equation. So on the left side, I&#x27;ll have why, over X plus one cubed it&#x27;s equal to. And then when I integrate the right side, get minus one over X plus one, plus a constant of integration and what the next this is, multiply everything by X plus one cube and so we end up with why is equal to X plus one squared plus c x plus, one cubed and one small correction. Here I forgot the minus sign eso that is my final answer and they were asked to go take a look and say graph this and see what different boundary conditions would happen. So looking at that constant of integration, so you see if that constant is zero, so you can see our any. It just Paul&#x27;s this. If the constant is zero, then this just becomes a parabola. So let&#x27;s sort of see what could make that happen. So constant becomes zero. So if that constant is zero Ah, come on. Here, Constant zero. I have a parabola minus X plus one squared opens down. And then as the constant changes, then it&#x27;s a polynomial. That&#x27;s a cubic So either gonna have a cubic polynomial or a quadratic or Cuba. She only had the quadratic at the initial condition that gives you see equal to zero. And then you see, this is what the family of solutions looks like for, uh, this thistle differential equation.

Problem

Solve the following differential equations. Use y…

Problem 236 Hard Difficulty

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?
$$
x y^{\prime}=2 \frac{\cos x}{x}-3 y
$$

Answer

$$y=\frac{2}{x^2} \sin x+\frac{2}{x^3} \cos x+\frac{C}{x^3}$$
The solutions change their behaviors on using an initial condition $y(x)$ approaching zero with small $x$ . In general, the solution behavior depend on the initial condition

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Gilbert Strang,

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$$y=\frac{2}{x^2} \sin x+\frac{2}{x^3} \cos x+\frac{C}{x^3}$$The solutions change their behaviors on using an initial condition $y(x)$ approaching zero with small $x$ . In general, the solution behavior depend on the initial condition

Calculus

Caleb E.

Kristen K.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Gilbert Strang,Calculus

Caleb E.

Kristen K.

Samuel H.

Joseph L.

$$y=\frac{2}{x^2} \sin x+\frac{2}{x^3} \cos x+\frac{C}{x^3}$$
The solutions change their behaviors on using an initial condition $y(x)$ approaching zero with small $x$ . In general, the solution behavior depend on the initial condition

Gilbert Strang,

Calculus