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In Exercises $1-14,$ find the general solution of the differential equation.

$$

\frac{d y}{d x}=\frac{x^{2}-3}{6 y^{2}}

$$

$y=\sqrt[3]{\frac{x^{3}}{6}} \frac{3 x}{2}+C$

Differential Equations

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Missouri State University

Campbell University

Oregon State University

Harvey Mudd College

hi, everyone. Let's go and solve For the general solution of the differential equation D y over D X is equal to y squared minus 3/6 y squared. Now, in order to find are the generals reminder that the general solution of a differential equation is the form where you can write one variables will pick why, as a function another variable which in this case would be X meaning that there is no other way. There's no nothing else that why depends on but X. So whatever X is is all you need to know in order to find out why is like when you were in sixth grade and you learn that why equals m x plus B for graphs that that's a differential equation. So in order to solve for this general form, we want to get the Y terms and the ex terms on the left and right side of the function, respectively. So in order to do this, we're just gonna use multiplication so on both sides were gonna multiply. Bye six y squared DX. No, doing this is okay because it's essentially like we're multiplying the entire equation by one in a sense where we're multiplying them both by the same thing. So it's okay, so on the left, all the X terms will cancel out and on the right, all the white terms will cancel out. And then we're just left with simply six y squared de y is equal to X squared minus three d x. So now we're just going to go ahead and solve the both sides the equation By integrating, we're gonna take the indefinite integral, so we don't need any bounds of each side. So we're here. We'll do in a role of six y squared with respect to why Over here you can you can split this into two different intervals if you want. That makes it easier for you. I'm going to leave them as one in this example. But how? I typed up the answer on like my text form of solution their separate. So it doesn't matter. So anyways, in a role of X squared minus three DX So now let's do the left hand side. So where you just simple interval just plus one to the exponents and then divide by the new exponents. So we got Sorry, we don't need that. So we simply just have plus one three and then divide by three. So we'll get two y cube and then on the other side we'll get so the X squared term will just be X squared, divided by three or sorry X cubed, divided by three hoops and then the term that doesn't have an excellent Well, so it's kind of like X to the zero power. So we'll add one to that. So we'll just be xto the first power, which is just X and then divided by one, which is still just x. So it will be minus three X. So anyways, oh, we also need to add on the right hand side we're gonna add a general constant, which was going to call, see, And that's because we don't have, uh, balance for inter goals. So we can't actually determine if there is a constant here or not, because when you would take the derivative of a constant, it would disappear. So with by us kind of undoing derivatives by doing intervals, we're adding this imaginary constant back into the equation. So, anyways, now that we have in this form, we just need to solve for? Why? Which is pretty straightforward. We're going to divide. We're gonna divide both sides by two first. Now we'll get why Cube equals two X cubed over three. Sorry about that. Or execute over. Yeah, execute over six minus three X over to and then we're still going to add Just see. Think of it this way. When hope they're, you know, Indians advisors right there when four divided by two is to before and two are both constants. So since we don't know what the constant C is, dividing it by another constant doesn't change mathematically what it iss. So that's why that's kind of useful. So now all we need to do is just take the cubic root and we'll get our answer. So we'll take the Cuban route of the whole term because again, we don't know what the sea is. So we have to include the whole term as one thing in are, um, square root sign or as in our cubic root term, Yeah. So why equals to keep a group of extra third minus six minus three x over two plus C. And that is your answer. And thank you for watching. And I hope this video helped you solve the problem.