Video: Solving a First-Order Separable Differential Equation Involving Integration by Parts and by Substitution

Video Transcript

Solve the differential equation the derivative of 𝜃 with respect to 𝑡 is equal to 𝑡 multiplied by the sec of 𝜃 divided by 𝜃 multiplied by 𝑒 to the power of 𝑡 squared.

The first thing we can try is to see if we’ve been given a separable differential equation. And we recall that we call a differential equation separable if the derivative of 𝜃 with respect to 𝑡 could be written as a function of 𝜃 multiplied by a function of 𝑡. And we see that we can separate our differential equation. We get the first function of the sec of 𝜃 divided by 𝜃. And then this is multiplied by 𝑡 divided by 𝑒 to the power of 𝑡 squared.

So since we have a separable differential equation, we’ll divide both sides of our equation by 𝑓 of 𝜃. And this is the same as multiplying by 𝜃 divided by the sec of 𝜃. This gives us 𝜃 divided by the sec of 𝜃 multiplied by the derivative of 𝜃 with respect to 𝑡 is equal to 𝑡 divided by 𝑒 to the power of 𝑡 squared.

It’s worth reiterating at this point that the derivative of 𝜃 with respect to 𝑡 is not a fraction. However, when we’re solving separable differential equations, we can treat it a little bit like a fraction. So using this gives us the equivalent statement that 𝜃 divided by the sec of 𝜃 d𝜃 is equal to 𝑡 divided by 𝑒 to the power of 𝑡 squared d𝑡.

Finally, we integrate both sides of this equation to get a new expression. We can simplify our integrand of 𝜃 divided by the sec of 𝜃 by recalling that the sec of 𝜃 is equal to one divided by the cos of 𝜃. And we can simplify the integrand in our second integral by recalling that dividing by 𝑒 to the power of 𝑡 squared is the same as multiplying by 𝑒 to the power of negative 𝑡 squared.

This gives us two integrals to solve. We can solve the first integral of 𝜃 multiplied by the cos of 𝜃 with respect to 𝜃 by using integration by parts. We recall that integration by parts tells us that the integral of 𝑢 multiplied by 𝑣 prime with respect to 𝜃 is equal to 𝑢 multiplied by 𝑣 minus the integral of 𝑢 prime multiplied by 𝑣 with respect to 𝜃.

So to evaluate the integral of 𝜃 multiplied by the cos of 𝜃 with respect to 𝜃, we’ll set 𝑢 equal to 𝜃 and 𝑣 prime equal to the cos of 𝜃. Differentiating 𝑢 with respect to 𝜃 gives us that 𝑢 prime is equal to one. And if we have that 𝑣 prime is equal to the cos of 𝜃, then our antiderivative 𝑣 is equal to the sin of 𝜃. So substituting this into our integration-by-parts formula gives us the integral of 𝜃 multiplied by the cos of 𝜃 with respect to 𝜃 is equal to 𝑢 multiplied by 𝑣, which is 𝜃 sin 𝜃, minus the integral of 𝑢 prime multiplied by 𝑣, which is one minus sin 𝜃, with respect to 𝜃.

Finally, we know that the integral of sin 𝜃 with respect to 𝜃 is equal to negative cos of 𝜃. So we’ve shown that the integral of 𝜃 cos of 𝜃 with respect to 𝜃 is equal to 𝜃 sin 𝜃 plus cos of 𝜃 plus our constant of integration 𝑐 one.

Now we’ve evaluated the first of our two integrals. This gives us that 𝜃 sin 𝜃 plus the cos of 𝜃 plus our constant of integration 𝑐 one is equal to the integral of 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared with respect to 𝑡.

Now we have to evaluate the integral of 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared with respect to 𝑡. And there are several ways of doing this. For example, we could use the substitution 𝑢 is equal to 𝑡 squared. However, there’s a slightly simpler way of doing this. If we were to try to differentiate 𝑒 to the power of negative 𝑡 squared with respect to 𝑡. Well, we recall by the chain rule the derivative of 𝑒 to the power of 𝑓 of 𝑥 with respect to 𝑥 is equal to 𝑓 prime of 𝑥 multiplied by 𝑒 to the power of 𝑓 of 𝑥.

So to differentiate 𝑒 to the power of negative 𝑡 squared with respect to 𝑡, we differentiate negative 𝑡 squared to get negative two 𝑡. And then we multiply this by 𝑒 to the power of negative 𝑡 squared. And we can now see that this is very similar to our integrand. In fact, it’s only different by a coefficient of negative a half. So this tells us that the derivative with respect to 𝑡 of negative a half multiplied by 𝑒 to the power of negative 𝑡 squared is equal to negative a half multiplied by negative two 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared.

We can then cancel the coefficient of negative a half with the coefficient of negative two to just get 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared. And we can see that this is exactly equal to our integrand.

So what we have shown is that negative a half multiplied by 𝑒 to of the power of negative 𝑡 squared is our antiderivative. So the integral of 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared with respect to 𝑡 is equal to negative a half multiplied by 𝑒 to the power of negative 𝑡 squared plus a constant of integration 𝑐 two. And since 𝑐 one and 𝑐 two are both constants, we can combine both of these into a new constant, which we will call 𝑐.

Therefore, we’ve shown that the solution to the differential equation given to us in the question is 𝜃 sin 𝜃 plus the cos of 𝜃 is equal to negative 𝑒 to the power of negative 𝑡 squared divided by two plus a constant of integration 𝑐.