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 ๐‘ฆ๐‘ฆโ€ฒ = (๐‘ฆ + 1)ยฒ, ๐‘ฆ โ‰  โˆ’1.

07:22

Video Transcript

Find the one-parameter family of solutions for the differential equation ๐‘ฆ๐‘ฆ prime is equal to ๐‘ฆ plus one all squared. ๐‘ฆ is not equal to negative one.

In this question, weโ€™re given the differential equation which weโ€™re asked to solve. And in fact, we need to find a family of solutions in one parameter. First, letโ€™s remember what a family of solutions in one parameter would mean. The most common way we wouldโ€™ve seen solutions like this is solutions of the form ๐‘ฆ is equal to ๐‘“ of ๐‘ฅ plus some constant ๐ถ, where this is a solution to our differential equation for any constant ๐ถ. This is an example of a solution in one parameter since this is a solution for any value of ๐ถ.

However, itโ€™s not always possible to write our solutions in this form. For example, we canโ€™t always find the solution of the form ๐‘ฆ is some function of ๐‘ฅ. But the important thing here to realize is the question is not asking us for a specific solution. Instead, itโ€™s asking us for a family of solutions.

Letโ€™s now move on to solving our differential equation. We know several different methods for solving differential equations. Weโ€™ll start by rewriting ๐‘ฆ prime as d๐‘ฆ by d๐‘ฅ. And itโ€™s worth pointing out here weโ€™re not told what ๐‘ฆ is a function of. Weโ€™ll call this variable ๐‘ฅ. However, we could call this whatever we want. The first method weโ€™ll try is solving this differential equation directly. We need to make d๐‘ฆ by d๐‘ฅ the subject. And we can do this by dividing both sides of our differential equation through by ๐‘ฆ. This gives us d๐‘ฆ by d๐‘ฅ is equal to ๐‘ฆ plus one all squared all divided by ๐‘ฆ.

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 ๐‘ฆ. This would then give us d๐‘ฆ by d๐‘ฅ is equal to ๐‘ฆ plus two plus one over ๐‘ฆ. However, this is not an easy differential equation to solve. Instead, letโ€™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.

Weโ€™ll start by multiplying both sides of our equation through by ๐‘ฆ and then dividing both sides of our questions through by ๐‘ฆ plus one all squared. This gives us ๐‘ฆ over ๐‘ฆ plus one all squared times d๐‘ฆ by d๐‘ฅ is equal to one. Now we need to rewrite this in terms of differentials. And when doing this, itโ€™s important to reiterate d๐‘ฆ by d๐‘ฅ is not a fraction. However, when weโ€™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 ๐‘ฆ over ๐‘ฆ plus one all squared d๐‘ฆ is equal to one d๐‘ฅ.

And itโ€™s also worth pointing out here some people like to simplify the right-hand side of this equation to just be d๐‘ฅ. Weโ€™ll leave this as one d๐‘ฅ. And to attempt to solve this differential equation, weโ€™re now going to integrate both sides. So we now have two indefinite integrals we need to solve. Letโ€™s start with the integral on the right-hand side of this equation, the integral of one with respect to ๐‘ฅ. We know how to do this by using the power rule for integration. We get ๐‘ฅ plus our constant of integration ๐ถ.

Now we want to evaluate the integral on the left-hand side of our equation. Thatโ€™s the integral of ๐‘ฆ over ๐‘ฆ plus one all squared with respect to ๐‘ฆ. 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 itโ€™s a standard method of evaluating integrals of this form, which we should be comfortable doing.

However, this is a long method, and thereโ€™s actually a simpler method. Recall that weโ€™re looking for solutions where ๐‘ฆ is not equal to negative one. If ๐‘ฆ is not allowed to be equal to negative one, this means we can cancel shared factors of ๐‘ฆ plus one because itโ€™s not equal to zero. And to see how we could do this, we could add one and subtract one from our numerator. This wonโ€™t change the value of our numerator. However, it does give us a factor of ๐‘ฆ plus one in our numerator. Weโ€™ll then split our integrand into two fractions, one with ๐‘ฆ 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 ๐‘ฆ plus one over ๐‘ฆ plus one all squared minus one over ๐‘ฆ plus one all squared with respect to ๐‘ฆ.

And since ๐‘ฆ is not equal to negative one, ๐‘ฆ plus one is not equal to zero. So we can cancel the shared factor of ๐‘ฆ plus one in our first term of the integrand. This means weโ€™ve rewritten the first term in our integrand as one over ๐‘ฆ plus one. In fact, this integrand is exactly what we wouldโ€™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 ๐‘ฆ plus one with respect to ๐‘ฆ minus the integral of one over ๐‘ฆ plus one all squared with respect to ๐‘ฆ. So now we have two integrals we need to evaluate. Letโ€™s start with our first integral. Thereโ€™s several different methods we could use to evaluate this integral. For example, we could use the substitution ๐‘ข is equal to ๐‘ฆ plus one.

And in fact, thereโ€™s two integral rules we can use to evaluate this integral. We know the integral of ๐‘“ prime of ๐‘ฆ divided by ๐‘“ of ๐‘ฆ with respect to ๐‘ฆ is equal to the natural logarithm of the absolute value of ๐‘“ of ๐‘ฆ plus a constant of integration ๐ถ. And this is valid so long as ๐‘“ of ๐‘ฆ is not equal to zero. In our case, this just means that ๐‘ฆ is not equal to negative one. Similarly, we can also use another integral rule, the integral of one over ๐‘ฆ plus ๐‘Ž with respect to ๐‘ฆ is equal to the natural logarithm of the absolute value of ๐‘ฆ plus ๐‘Ž plus a constant of integration ๐ถ. And this is true for any constant ๐‘Ž and as long as ๐‘ฆ plus ๐‘Ž is not equal to zero. In other words, ๐‘ฆ 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 ๐‘ฆ plus one. And itโ€™s worth pointing out we donโ€™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. Weโ€™ll do this by using the substitution ๐‘ข is equal to ๐‘ฆ plus one. So using the substitution ๐‘ข is equal to ๐‘ฆ plus one, weโ€™ll differentiate both sides with respect to ๐‘ฆ. We get d๐‘ข by d๐‘ฆ is equal to one. And this gives us the equivalent statement in terms of differentials, d๐‘ข is equal to d๐‘ฆ. So by using the substitution ๐‘ข is equal to ๐‘ฆ plus one, we need to subtract the integral of one over ๐‘ข squared with respect to ๐‘ข. And we can evaluate this integral by using the power rule for integration. It might help us to think of our integrand as ๐‘ข to the power of negative two.

We want to add one to our exponent of ๐‘ข and then divide by this new exponent. This gives us ๐‘ข to the power of negative one divided by negative one. And remember, weโ€™re subtracting this integral. And this will be true for all values of ๐‘ข except where ๐‘ข is equal to zero. But ๐‘ข being equal to zero means that ๐‘ฆ is equal to negative one. And weโ€™re only looking for solutions where ๐‘ฆ is not equal to negative one. So we can use this and then simplify. First, weโ€™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 ๐‘ข in the denominator. And then we can use our substitution ๐‘ข is equal to ๐‘ฆ plus one.

This gives us the natural logarithm of the absolute value of ๐‘ฆ plus one plus one over ๐‘ฆ plus one. And of course, we could add constants of integrations for all of these; however, weโ€™re just going to combine these at the end anyway. Weโ€™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 ๐‘ฆ plus one plus the natural logarithm of the absolute value of ๐‘ฆ plus one is equal to ๐‘ฅ plus ๐ถ. And this would be a solution to our differential equation for any value of ๐ถ. So this is a family of solutions in one parameter.

Therefore, given the differential equation ๐‘ฆ๐‘ฆ prime is equal to ๐‘ฆ plus one all squared where ๐‘ฆ is not equal to negative one, we were able to find the following family of solutions in one parameter. One over ๐‘ฆ plus one plus the natural logarithm of the absolute value of ๐‘ฆ plus one is equal to ๐‘ฅ plus ๐ถ.

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