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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?$$x y^{\prime}=2 \frac{\cos x}{x}-3 y$$

$$y=\frac{2}{x^2} \sin x+\frac{2}{x^3} \cos x+\frac{C}{x^3}$$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 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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Section four, not five. Problem to 36. We'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'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'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'm gonna end up with X cubed. Why is equal to when I integrate this? I'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's, ah, change. When c is equal to zero, that'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

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