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

Solve the differential equation dπœƒ/d𝑑 = (𝑑 sec πœƒ)/(πœƒπ‘’^(𝑑²)).

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Video Transcript

Solve the differential equation the derivative of πœƒ with respect to 𝑑 is equal to 𝑑 multiplied by the sec of πœƒ divided by πœƒ multiplied by 𝑒 to the power of 𝑑 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 πœƒ with respect to 𝑑 could be written as a function of πœƒ multiplied by a function of 𝑑. And we see that we can separate our differential equation. We get the first function of the sec of πœƒ divided by πœƒ. And then this is multiplied by 𝑑 divided by 𝑒 to the power of 𝑑 squared.

So since we have a separable differential equation, we’ll divide both sides of our equation by 𝑓 of πœƒ. And this is the same as multiplying by πœƒ divided by the sec of πœƒ. This gives us πœƒ divided by the sec of πœƒ multiplied by the derivative of πœƒ with respect to 𝑑 is equal to 𝑑 divided by 𝑒 to the power of 𝑑 squared.

It’s worth reiterating at this point that the derivative of πœƒ with respect to 𝑑 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 πœƒ divided by the sec of πœƒ dπœƒ is equal to 𝑑 divided by 𝑒 to the power of 𝑑 squared d𝑑.

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

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

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

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

Now we’ve evaluated the first of our two integrals. This gives us that πœƒ sin πœƒ plus the cos of πœƒ plus our constant of integration 𝑐 one is equal to the integral of 𝑑 multiplied by 𝑒 to the power of negative 𝑑 squared with respect to 𝑑.

Now we have to evaluate the integral of 𝑑 multiplied by 𝑒 to the power of negative 𝑑 squared with respect to 𝑑. And there are several ways of doing this. For example, we could use the substitution 𝑒 is equal to 𝑑 squared. However, there’s a slightly simpler way of doing this. If we were to try to differentiate 𝑒 to the power of negative 𝑑 squared with respect to 𝑑. Well, we recall by the chain rule the derivative of 𝑒 to the power of 𝑓 of π‘₯ with respect to π‘₯ is equal to 𝑓 prime of π‘₯ multiplied by 𝑒 to the power of 𝑓 of π‘₯.

So to differentiate 𝑒 to the power of negative 𝑑 squared with respect to 𝑑, we differentiate negative 𝑑 squared to get negative two 𝑑. And then we multiply this by 𝑒 to the power of negative 𝑑 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 𝑑 of negative a half multiplied by 𝑒 to the power of negative 𝑑 squared is equal to negative a half multiplied by negative two 𝑑 multiplied by 𝑒 to the power of negative 𝑑 squared.

We can then cancel the coefficient of negative a half with the coefficient of negative two to just get 𝑑 multiplied by 𝑒 to the power of negative 𝑑 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 𝑑 squared is our antiderivative. So the integral of 𝑑 multiplied by 𝑒 to the power of negative 𝑑 squared with respect to 𝑑 is equal to negative a half multiplied by 𝑒 to the power of negative 𝑑 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 πœƒ sin πœƒ plus the cos of πœƒ is equal to negative 𝑒 to the power of negative 𝑑 squared divided by two plus a constant of integration 𝑐.

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