Question Video: Solving a First-Order Separable Differential Equation Involving Factorization by Grouping to Separate the Variables Mathematics • Higher Education

Video Transcript

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

The first thing we can try and do when we’re asked to solve a differential equation is try to write it as a separable differential equation. Remember, this is when it’s of the form d𝑝 by d𝑡 is equal to some function 𝑓of 𝑝 multiplied by some function 𝑔 of 𝑡. We can then attempt to solve this by separating our variables and using our integration techniques. So, let’s try and write the differential equation given to us in the question as a separable differential equation.

Initially, it doesn’t seem obvious why this would be the case. However, this doesn’t stop us attempting to do this. Let’s start by factoring all terms of our 𝑝 to try and find our function 𝑓 of 𝑝. Only our first two terms contain 𝑝. So, when we factor this out, we’ll get 𝑝 times 𝑡 squared minus five. So, we’ve rewritten our differential equation as d𝑝 by d𝑡 is equal to 𝑝 times 𝑡 squared minus five plus 𝑡 squared minus five.

And now, we can see our first term contains a factor of 𝑡 squared minus five and our second and third term combine to make 𝑡 squared minus five. So, let’s take out a factor of 𝑡 squared minus five. This gives us 𝑡 squared minus five multiplied by 𝑝 plus one. So, now, we can see we’ve written our differential equation as a separable differential equation. Our function 𝑓 of 𝑝 is 𝑝 plus one and our function 𝑔 of 𝑡 is 𝑡 squared minus five.

Now, to solve our separable differential equation, we’re going to want to separate our variables. We’ll do this by dividing both sides of our equation by 𝑝 plus one. Doing this and then simplifying, we get one over 𝑝 plus one d𝑝 by d𝑡 is equal to 𝑡 squared minus five. And remember, d𝑝 by d𝑡 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 one over 𝑝 plus one d𝑝 is equal to 𝑡 squared minus five d𝑡.

Finally, we’ll 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 𝑎, the integral of one divided by 𝑥 plus 𝑎 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑥 plus 𝑎 plus the constant of integration 𝐶. So, by using this rule with our variable called 𝑝 and our value of 𝑎 equal to one, we get the integral of one over 𝑝 plus one with respect to 𝑝 is equal to the natural logarithm of the absolute value of 𝑝 plus one plus the constant of integration we’ll call 𝐶 one.

So, we’ve evaluated our first integral. We now need to evaluate our second integral. That’s the integral of 𝑡 squared minus five with respect to 𝑡. We can do this directly by using the power rule for integration. We want to add one to our exponent of 𝑡 and then divide by this exponent of 𝑡. We get 𝑡 cubed over three minus five 𝑡 plus the constant of integration we’ve called 𝐶 two.

We can then simplify this expression. We know that 𝐶 one and 𝐶 two are both constants of integration. So, we’ll combine both of these into a new constant we’ll call 𝐶. So, this gives us the natural logarithm of the absolute value of 𝑝 plus one is equal to 𝑡 cubed over three minus five 𝑡 plus 𝐶.

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 𝑝 is some function of 𝑡. And in this case, it’s possible to do that. We’ll do this by writing 𝑒 to the power of both sides of our equation. This gives us 𝑒 to the power of the natural logarithm of the absolute value of 𝑝 plus one is equal to 𝑒 to the power of 𝑡 cubed over three minus five 𝑡 plus 𝐶.

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

Next, by using our laws of exponents, we’ll rewrite the right-hand side of our equation as 𝑒 to the power of 𝐶 times 𝑒 to the power of 𝑡 cubed over three minus five 𝑡. And we can simplify this expression even further. Since 𝐶 is a constant, 𝑒 to the power of 𝐶 is also a constant. So, we’ll just write this as some new constant we’ll call 𝐷. We now have the absolute value of 𝑝 plus one is equal to 𝐷 times 𝑒 to the power of 𝑡 cubed over three minus five 𝑡.

Remember, we’re trying to write this in the form 𝑝 is equal to some function of 𝑡. Now, we recall when we’re trying to solve the equation the absolute value of 𝑥 is equal to 𝑦, this is the same as solving the equation plus or minus 𝑥 is equal to 𝑦. So, solving the absolute value of 𝑝 plus one is equal to some function of 𝑡 is the same as solving the equation plus or minus 𝑝 plus one is equal to that function of 𝑡. We can then just multiply both sides of this equation through by negative one.

Remember, this means we’ll swap the sign of our positive and negative. But now, we see we have negative or positive 𝐷. But 𝐷 is just a constant. So, we can write this as some new constant we’ll call 𝐾. Finally, to write this in terms of 𝑝, we just need to subtract one from both sides of our equation. And this gives us our final general solution in the form 𝑝 is some function of 𝑡.

Therefore, we were able to solve the differential equation d𝑝 by d𝑡 is equal to 𝑡 squared 𝑝 minus five 𝑝 plus 𝑡 squared minus five. We found the general solution 𝑝 is equal to 𝐾 times 𝑒 to the power of one-third 𝑡 cubed minus five 𝑡 minus one.