Question Video: Solving a Separable First-Order Differential Equation Involving Integration by Substitution given Its Initial Condition Mathematics

Video Transcript

Find the solution of the differential equation 𝑥 plus 𝑦 squared times the square root of 𝑥 squared plus three times d𝑦 by d𝑥 is equal to zero that satisfies the initial condition 𝑦 of negative one is equal to negative two.

We’re given a differential equation, and we were asked to find the solution to this differential equation which satisfies the initial condition 𝑦 of negative one is equal to negative two. To do this, we’re first going to need to find a general solution to our differential equation. And there’s several different ways we could try to do this. In this case, we can see our solution is given in terms of 𝑥 and 𝑦, and we can’t simplify this any further. So, we could try writing this as a separable differential equation. So, we want to separate our variables 𝑥 and 𝑦 onto opposite sides of the equation. We’ll start by subtracting 𝑥 from both sides of our differential equation. Doing this, we get 𝑦 squared times the square root of 𝑥 squared plus three multiplied by d𝑦 by d𝑥 is equal to negative 𝑥.

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

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. Let’s start with the integral on the left-hand side, the integral of 𝑦 squared with respect to 𝑦. 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 𝑦 cubed over three. And we could add a constant of integration. However, we’re integrating on both sides of our equations. We only need to add one constant of integration. We’ll do this in our second integral.

Next, we need to evaluate the integral of negative 𝑥 divided by the square root of 𝑥 squared plus three. There’s a few different ways of doing this; however, the easiest is to use the substitution 𝑢 is equal to 𝑥 squared plus three. To integrate using a substitution, we need to find an expression for d𝑢 by d𝑥. That’s the derivative of 𝑥 squared plus three with respect to 𝑥. We can find this term by term by using the power rule for differentiation. We get d𝑢 by d𝑥 is equal to two 𝑥. Once again, we do know that d𝑢 by d𝑥 is not a fraction.

However, when we’re 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, d𝑢 is equal to two 𝑥 d𝑥. We can see the two 𝑥 d𝑥 does not appear in our integrand. However, negative 𝑥 d𝑥 does appear in our integrand. So, it can help to rewrite this expression in terms of negative 𝑥 d𝑥. To do this, we need to divide both sides of our equation through by negative two. This gives us that negative one-half d𝑢 is equal to negative 𝑥 d𝑥.

We’re now ready to evaluate the integral of negative 𝑥 over the square root of 𝑥 squared plus three with respect to 𝑥 by using our substitution. First, we’ll replace negative 𝑥 d𝑥 with negative one-half d𝑢. Next, in our denominator, we’ll replace 𝑥 squared plus three with 𝑢. This gives us the integral of negative one-half divided by the square root of 𝑢 with respect to 𝑢. All we need to do now is evaluate this integral. First, we’ll take the factor of negative one-half outside of our integral. Now, to integrate this, we’re going to want to use the power rule for integration. To do this, we’re going to need to recall by using our laws of exponents, dividing by the square root of 𝑢 is the same as multiplying by 𝑢 to the power of negative one-half. So, this gives us negative one-half multiplied by the integral of 𝑢 to the power of negative one-half with respect to 𝑢. 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. We’ll call this constant 𝐴. 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 𝑢 to the power of one-half plus 𝐴. And remember, this is equal to 𝑦 cubed divided by three. We can simplify this even further. 𝑢 to the power of one-half is equal to the square root of 𝑢. But remember, we’re trying to find a solution to a differential equation in terms of 𝑦 and 𝑥. So, we want our solution is going to be in terms of 𝑦 and 𝑥. So, we should use our substitution 𝑢 is equal to 𝑥 squared plus three. This gives us 𝑦 cubed over three is equal to negative the square root of 𝑥 squared plus three plus the constant of integration 𝐴. And this gives us the general solutions to our differential equation.

However, if it’s possible, we should try and write our solution 𝑦 as some function in 𝑥. And in this case, it is possible. First, we’re going to multiply both sides of our equation through by three. So, we get that 𝑦 cubed is equal to the following expression. We need to distribute three over our parentheses. Doing this gives us that 𝑦 cubed is equal to negative three times the square root of 𝑥 squared plus three plus three 𝐴. However, 𝐴 is just a constant. So, three 𝐴 will also be a constant. So, we can call this a new constant we’ll call 𝐶.

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

Substituting 𝑥 is equal to negative one and 𝑦 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 𝐶. Now, all that’s left to do is solve this equation for 𝐶. 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 𝐶. 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 𝐶. We’ll do this by cubing both sides of our equation. This gives us that negative eight is equal to negative six plus 𝐶. Finally, all we do is add six to both sides of this equation to see that 𝐶 is equal to negative two. So, now, we just substitute 𝐶 is equal to negative two into the general solution for our differential equation. This gives us that 𝑦 is equal to the cube root of negative three times the square root of 𝑥 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 𝑥 plus 𝑦 squared times the square root of 𝑥 squared plus three d𝑦 by d𝑥 is equal to zero that satisfies the initial condition 𝑦 of negative one is equal to negative two. This solution is 𝑦 is equal to negative the cube root of three times the square root of 𝑥 squared plus three plus two.