### Video Transcript

Find the solution of the differential equation ๐ฅ plus ๐ฆ squared times the square root of ๐ฅ squared plus three times d๐ฆ by d๐ฅ is equal to zero that satisfies the initial condition ๐ฆ of negative one is equal to negative two.

Weโre given a differential equation, and we were asked to find the solution to this differential equation which satisfies the initial condition ๐ฆ of negative one is equal to negative two. To do this, weโre first going to need to find a general solution to our differential equation. And thereโs several different ways we could try to do this. In this case, we can see our solution is given in terms of ๐ฅ and ๐ฆ, and we canโt simplify this any further. So, we could try writing this as a separable differential equation. So, we want to separate our variables ๐ฅ and ๐ฆ onto opposite sides of the equation. Weโll start by subtracting ๐ฅ from both sides of our differential equation. Doing this, we get ๐ฆ squared times the square root of ๐ฅ squared plus three multiplied by d๐ฆ by d๐ฅ is equal to negative ๐ฅ.

However, weโre still not done. Weโre going to need to divide both sides of our equation through by the square root of ๐ฅ squared plus three. So, by dividing through by this, we get ๐ฆ squared times d๐ฆ by d๐ฅ would be equal to negative ๐ฅ divided by the square root of ๐ฅ squared plus three. And now, we can see this is a separable differential equation. And remember, d๐ฆ by d๐ฅ is not a fraction. However, when weโre solving separable differential equations, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials. ๐ฆ squared d๐ฆ is equal to negative ๐ฅ over the square root of ๐ฅ squared plus three d๐ฅ.

And all we need to do now is integrate both sides of this equation. This will help us find our solution to our original differential equation. We need to evaluate both of these integrals. Letโs start with the integral on the left-hand side, the integral of ๐ฆ squared with respect to ๐ฆ. We can do this by using the power rule for integration. We want to add one to our exponent of two, giving us a new exponent of three, and then divide by this new exponent. This gives us ๐ฆ cubed over three. And we could add a constant of integration. However, weโre integrating on both sides of our equations. We only need to add one constant of integration. Weโll do this in our second integral.

Next, we need to evaluate the integral of negative ๐ฅ divided by the square root of ๐ฅ squared plus three. Thereโs a few different ways of doing this; however, the easiest is to use the substitution ๐ข is equal to ๐ฅ squared plus three. To integrate using a substitution, we need to find an expression for d๐ข by d๐ฅ. Thatโs the derivative of ๐ฅ squared plus three with respect to ๐ฅ. We can find this term by term by using the power rule for differentiation. We get d๐ข by d๐ฅ is equal to two ๐ฅ. Once again, we do know that d๐ข by d๐ฅ is not a fraction.

However, when weโre using an integration by substitution, it can help to think of it a little bit like a fraction. This gives us the equivalent statement in terms of differentials, d๐ข is equal to two ๐ฅ d๐ฅ. We can see the two ๐ฅ d๐ฅ does not appear in our integrand. However, negative ๐ฅ d๐ฅ does appear in our integrand. So, it can help to rewrite this expression in terms of negative ๐ฅ d๐ฅ. To do this, we need to divide both sides of our equation through by negative two. This gives us that negative one-half d๐ข is equal to negative ๐ฅ d๐ฅ.

Weโre now ready to evaluate the integral of negative ๐ฅ over the square root of ๐ฅ squared plus three with respect to ๐ฅ by using our substitution. First, weโll replace negative ๐ฅ d๐ฅ with negative one-half d๐ข. Next, in our denominator, weโll replace ๐ฅ squared plus three with ๐ข. This gives us the integral of negative one-half divided by the square root of ๐ข with respect to ๐ข. All we need to do now is evaluate this integral. First, weโll take the factor of negative one-half outside of our integral. Now, to integrate this, weโre going to want to use the power rule for integration. To do this, weโre going to need to recall by using our laws of exponents, dividing by the square root of ๐ข is the same as multiplying by ๐ข to the power of negative one-half. So, this gives us negative one-half multiplied by the integral of ๐ข to the power of negative one-half with respect to ๐ข. And we can evaluate this integral by using the power rule for integration. We add one to our exponent giving us a new exponent of one-half and then divide by the new exponent of one-half.

And remember, we need to add a constant of integration. Because this is a constant, we can do this right at the end of our expression. Weโll call this constant ๐ด. And we can simplify this expression. We can cancel the shared factor of one-half in our numerator and our denominator. This gives us negative one times ๐ข to the power of one-half plus ๐ด. And remember, this is equal to ๐ฆ cubed divided by three. We can simplify this even further. ๐ข to the power of one-half is equal to the square root of ๐ข. But remember, weโre trying to find a solution to a differential equation in terms of ๐ฆ and ๐ฅ. So, we want our solution is going to be in terms of ๐ฆ and ๐ฅ. So, we should use our substitution ๐ข is equal to ๐ฅ squared plus three. This gives us ๐ฆ cubed over three is equal to negative the square root of ๐ฅ squared plus three plus the constant of integration ๐ด. And this gives us the general solutions to our differential equation.

However, if itโs possible, we should try and write our solution ๐ฆ as some function in ๐ฅ. And in this case, it is possible. First, weโre going to multiply both sides of our equation through by three. So, we get that ๐ฆ cubed is equal to the following expression. We need to distribute three over our parentheses. Doing this gives us that ๐ฆ cubed is equal to negative three times the square root of ๐ฅ squared plus three plus three ๐ด. However, ๐ด is just a constant. So, three ๐ด will also be a constant. So, we can call this a new constant weโll call ๐ถ.

Finally, all we need to do is take the cube roots of both sides of this equation. And doing this, we get that ๐ฆ is equal to the cube root of negative three times the square root of ๐ฅ squared plus three plus the constant of integration ๐ถ. And this is the general solution to our differential equation. However, the question wants us to find the specific solution where ๐ฆ of negative one is equal to two. In other words, when ๐ฅ is equal to negative one, ๐ฆ is equal to negative two. We need to substitute these into our general solution to our differential equation to find the value of the constant ๐ถ.

Substituting ๐ฅ is equal to negative one and ๐ฆ is equal to negative two into our general solution to our differential equation gives us that negative two should be equal to the cube root of negative three times the square root of negative one squared plus three plus the constant of integration ๐ถ. Now, all thatโs left to do is solve this equation for ๐ถ. First, negative one squared plus three will be equal to four. And, of course, the square root of four is going to be equal to two. So, we have that negative two should be equal to the cube root of negative three times two plus the constant of integration ๐ถ. Of course, we can simplify this further. Negative three multiplied by two is equal the negative six.

So now, we need to solve negative two is equal to the cube root of negative six plus ๐ถ. Weโll do this by cubing both sides of our equation. This gives us that negative eight is equal to negative six plus ๐ถ. Finally, all we do is add six to both sides of this equation to see that ๐ถ is equal to negative two. So, now, we just substitute ๐ถ is equal to negative two into the general solution for our differential equation. This gives us that ๐ฆ is equal to the cube root of negative three times the square root of ๐ฅ squared plus three minus two. We could leave our answer like this. However, we can simplify this slightly. We have a factor of negative one inside of our cube root, but the cube root of negative one is just equal to negative one. So, we can take this outside of our cube root. And this gives us our final answer.

Therefore, we were able to find the solution of the differential equation ๐ฅ plus ๐ฆ squared times the square root of ๐ฅ squared plus three d๐ฆ by d๐ฅ is equal to zero that satisfies the initial condition ๐ฆ of negative one is equal to negative two. This solution is ๐ฆ is equal to negative the cube root of three times the square root of ๐ฅ squared plus three plus two.