# Question Video: Finding a 1-Parameter Family of Solutions to a Given Differential Equation

Find the 1-parameter family of solutions for the differential equation 𝑦𝑦′ = (𝑦 + 1)², 𝑦 ≠ −1.

07:22

### Video Transcript

Find the one-parameter family of solutions for the differential equation 𝑦𝑦 prime is equal to 𝑦 plus one all squared. 𝑦 is not equal to negative one.

In this question, we’re given the differential equation which we’re asked to solve. And in fact, we need to find a family of solutions in one parameter. First, let’s remember what a family of solutions in one parameter would mean. The most common way we would’ve seen solutions like this is solutions of the form 𝑦 is equal to 𝑓 of 𝑥 plus some constant 𝐶, where this is a solution to our differential equation for any constant 𝐶. This is an example of a solution in one parameter since this is a solution for any value of 𝐶.

However, it’s not always possible to write our solutions in this form. For example, we can’t always find the solution of the form 𝑦 is some function of 𝑥. But the important thing here to realize is the question is not asking us for a specific solution. Instead, it’s asking us for a family of solutions.

Let’s now move on to solving our differential equation. We know several different methods for solving differential equations. We’ll start by rewriting 𝑦 prime as d𝑦 by d𝑥. And it’s worth pointing out here we’re not told what 𝑦 is a function of. We’ll call this variable 𝑥. However, we could call this whatever we want. The first method we’ll try is solving this differential equation directly. We need to make d𝑦 by d𝑥 the subject. And we can do this by dividing both sides of our differential equation through by 𝑦. This gives us d𝑦 by d𝑥 is equal to 𝑦 plus one all squared all divided by 𝑦.

And we could try and solve this differential equation directly. For example, we could distribute the square over our parentheses and then divide through by 𝑦. This would then give us d𝑦 by d𝑥 is equal to 𝑦 plus two plus one over 𝑦. However, this is not an easy differential equation to solve. Instead, let’s go back to the previous step. We can actually notice at this point this is a separable differential equation. So we can try solving this by separating our variables onto opposite sides of the equation.

We’ll start by multiplying both sides of our equation through by 𝑦 and then dividing both sides of our questions through by 𝑦 plus one all squared. This gives us 𝑦 over 𝑦 plus one all squared times d𝑦 by d𝑥 is equal to one. Now we need to rewrite this in terms of differentials. And when doing this, it’s important to reiterate 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 over 𝑦 plus one all squared d𝑦 is equal to one d𝑥.

And it’s also worth pointing out here some people like to simplify the right-hand side of this equation to just be d𝑥. We’ll leave this as one d𝑥. And to attempt to solve this differential equation, we’re now going to integrate both sides. So we now have two indefinite integrals we need to solve. Let’s start with the integral on the right-hand side of this equation, the integral of one with respect to 𝑥. We know how to do this by using the power rule for integration. We get 𝑥 plus our constant of integration 𝐶.

Now we want to evaluate the integral on the left-hand side of our equation. That’s the integral of 𝑦 over 𝑦 plus one all squared with respect to 𝑦. And there are a lot of different ways of evaluating this integral. One way to do this is by using partial fractions to simplify our integrand. And this would work. And it’s a standard method of evaluating integrals of this form, which we should be comfortable doing.

However, this is a long method, and there’s actually a simpler method. Recall that we’re looking for solutions where 𝑦 is not equal to negative one. If 𝑦 is not allowed to be equal to negative one, this means we can cancel shared factors of 𝑦 plus one because it’s not equal to zero. And to see how we could do this, we could add one and subtract one from our numerator. This won’t change the value of our numerator. However, it does give us a factor of 𝑦 plus one in our numerator. We’ll then split our integrand into two fractions, one with 𝑦 plus one in the numerator and the other one with negative one in the numerator. This means we can rewrite this integral as the integral of 𝑦 plus one over 𝑦 plus one all squared minus one over 𝑦 plus one all squared with respect to 𝑦.

And since 𝑦 is not equal to negative one, 𝑦 plus one is not equal to zero. So we can cancel the shared factor of 𝑦 plus one in our first term of the integrand. This means we’ve rewritten the first term in our integrand as one over 𝑦 plus one. In fact, this integrand is exactly what we would’ve gotten if we had applied partial fractions.

We now want to integrate each term in our integrand separately. This gives us the integral of one over 𝑦 plus one with respect to 𝑦 minus the integral of one over 𝑦 plus one all squared with respect to 𝑦. So now we have two integrals we need to evaluate. Let’s start with our first integral. There’s several different methods we could use to evaluate this integral. For example, we could use the substitution 𝑢 is equal to 𝑦 plus one.

And in fact, there’s two integral rules we can use to evaluate this integral. We know the integral of 𝑓 prime of 𝑦 divided by 𝑓 of 𝑦 with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑦 plus a constant of integration 𝐶. And this is valid so long as 𝑓 of 𝑦 is not equal to zero. In our case, this just means that 𝑦 is not equal to negative one. Similarly, we can also use another integral rule, the integral of one over 𝑦 plus 𝑎 with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑦 plus 𝑎 plus a constant of integration 𝐶. And this is true for any constant 𝑎 and as long as 𝑦 plus 𝑎 is not equal to zero. In other words, 𝑦 is not allowed to be equal to negative one.

Using any of these methods, we get this integral evaluates to give us the natural logarithm of the absolute value of 𝑦 plus one. And it’s worth pointing out we don’t need to include a constant of integration because we can combine all of our constant of integrations into the one we already have.

Now we want to evaluate our second integral. We’ll do this by using the substitution 𝑢 is equal to 𝑦 plus one. So using the substitution 𝑢 is equal to 𝑦 plus one, we’ll differentiate both sides with respect to 𝑦. We get d𝑢 by d𝑦 is equal to one. And this gives us the equivalent statement in terms of differentials, d𝑢 is equal to d𝑦. So by using the substitution 𝑢 is equal to 𝑦 plus one, we need to subtract the integral of one over 𝑢 squared with respect to 𝑢. And we can evaluate this integral by using the power rule for integration. It might help us to think of our integrand as 𝑢 to the power of negative two.

We want to add one to our exponent of 𝑢 and then divide by this new exponent. This gives us 𝑢 to the power of negative one divided by negative one. And remember, we’re subtracting this integral. And this will be true for all values of 𝑢 except where 𝑢 is equal to zero. But 𝑢 being equal to zero means that 𝑦 is equal to negative one. And we’re only looking for solutions where 𝑦 is not equal to negative one. So we can use this and then simplify. First, we’re subtracting something divided by negative one. So instead, we can just add this and cancel the negative one. Next, by using our laws of exponents, we can write 𝑢 in the denominator. And then we can use our substitution 𝑢 is equal to 𝑦 plus one.

This gives us the natural logarithm of the absolute value of 𝑦 plus one plus one over 𝑦 plus one. And of course, we could add constants of integrations for all of these; however, we’re just going to combine these at the end anyway. We’ve now integrated both of our expressions, so we can just we can write these in. Writing both of our expressions for these and then combining our constants on the right-hand side, we get one over 𝑦 plus one plus the natural logarithm of the absolute value of 𝑦 plus one is equal to 𝑥 plus 𝐶. And this would be a solution to our differential equation for any value of 𝐶. So this is a family of solutions in one parameter.

Therefore, given the differential equation 𝑦𝑦 prime is equal to 𝑦 plus one all squared where 𝑦 is not equal to negative one, we were able to find the following family of solutions in one parameter. One over 𝑦 plus one plus the natural logarithm of the absolute value of 𝑦 plus one is equal to 𝑥 plus 𝐶.