### Video Transcript

Solve the following differential equation: the derivative of π with respect to π‘ is equal to π‘ squared π minus five π plus π‘ squared minus five.

The first thing we can try and do when weβre asked to solve a differential equation is try to write it as a separable differential equation. Remember, this is when itβs of the form dπ by dπ‘ is equal to some function πof π multiplied by some function π of π‘. We can then attempt to solve this by separating our variables and using our integration techniques. So, letβs try and write the differential equation given to us in the question as a separable differential equation.

Initially, it doesnβt seem obvious why this would be the case. However, this doesnβt stop us attempting to do this. Letβs start by factoring all terms of our π to try and find our function π of π. Only our first two terms contain π. So, when we factor this out, weβll get π times π‘ squared minus five. So, weβve rewritten our differential equation as dπ by dπ‘ is equal to π times π‘ squared minus five plus π‘ squared minus five.

And now, we can see our first term contains a factor of π‘ squared minus five and our second and third term combine to make π‘ squared minus five. So, letβs take out a factor of π‘ squared minus five. This gives us π‘ squared minus five multiplied by π plus one. So, now, we can see weβve written our differential equation as a separable differential equation. Our function π of π is π plus one and our function π of π‘ is π‘ squared minus five.

Now, to solve our separable differential equation, weβre going to want to separate our variables. Weβll do this by dividing both sides of our equation by π plus one. Doing this and then simplifying, we get one over π plus one dπ by dπ‘ is equal to π‘ squared minus five. And remember, dπ by dπ‘ is not a fraction. However, when weβre solving separable differential equations, we can treat it a little bit like a fraction. This gives us one over π plus one dπ is equal to π‘ squared minus five dπ‘.

Finally, weβll integrate both sides of this equation. So now, to solve our differential equation, we just need to evaluate both of these integrals. To start, we know for any real constant π, the integral of one divided by π₯ plus π with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus π plus the constant of integration πΆ. So, by using this rule with our variable called π and our value of π equal to one, we get the integral of one over π plus one with respect to π is equal to the natural logarithm of the absolute value of π plus one plus the constant of integration weβll call πΆ one.

So, weβve evaluated our first integral. We now need to evaluate our second integral. Thatβs the integral of π‘ squared minus five with respect to π‘. We can do this directly by using the power rule for integration. We want to add one to our exponent of π‘ and then divide by this exponent of π‘. We get π‘ cubed over three minus five π‘ plus the constant of integration weβve called πΆ two.

We can then simplify this expression. We know that πΆ one and πΆ two are both constants of integration. So, weβll combine both of these into a new constant weβll call πΆ. So, this gives us the natural logarithm of the absolute value of π plus one is equal to π‘ cubed over three minus five π‘ plus πΆ.

And this is the general solution to our differential equation. We could stop here and leave our answer like this. However, we should try and write our answer in the form π is some function of π‘. And in this case, itβs possible to do that. Weβll do this by writing π to the power of both sides of our equation. This gives us π to the power of the natural logarithm of the absolute value of π plus one is equal to π to the power of π‘ cubed over three minus five π‘ plus πΆ.

Using our laws of logarithms, we have π to the power of the natural logarithm of π is just equal to π. So, π to the power of the natural logarithm of the absolute value of π plus one is just the absolute value of π plus one.

Next, by using our laws of exponents, weβll rewrite the right-hand side of our equation as π to the power of πΆ times π to the power of π‘ cubed over three minus five π‘. And we can simplify this expression even further. Since πΆ is a constant, π to the power of πΆ is also a constant. So, weβll just write this as some new constant weβll call π·. We now have the absolute value of π plus one is equal to π· times π to the power of π‘ cubed over three minus five π‘.

Remember, weβre trying to write this in the form π is equal to some function of π‘. Now, we recall when weβre trying to solve the equation the absolute value of π₯ is equal to π¦, this is the same as solving the equation plus or minus π₯ is equal to π¦. So, solving the absolute value of π plus one is equal to some function of π‘ is the same as solving the equation plus or minus π plus one is equal to that function of π‘. We can then just multiply both sides of this equation through by negative one.

Remember, this means weβll swap the sign of our positive and negative. But now, we see we have negative or positive π·. But π· is just a constant. So, we can write this as some new constant weβll call πΎ. Finally, to write this in terms of π, we just need to subtract one from both sides of our equation. And this gives us our final general solution in the form π is some function of π‘.

Therefore, we were able to solve the differential equation dπ by dπ‘ is equal to π‘ squared π minus five π plus π‘ squared minus five. We found the general solution π is equal to πΎ times π to the power of one-third π‘ cubed minus five π‘ minus one.