Video: Solve an Initial Value Problem Using Inverse Trigonometic Functions

Solve the differential equation (d𝑦/dπ‘₯) (π‘₯Β² + 4) = 3 for y given that 𝑦(2)= 0.

03:27

Video Transcript

Solve the differential equation d𝑦 by dπ‘₯ multiplied by π‘₯ squared plus four is equal to three given that 𝑦 of two is equal to zero.

We’re tasked to solve a differential equation. And we can immediately see that this is a simple differential equation. We can divide both sides of this equation by π‘₯ squared plus four to get d𝑦 by dπ‘₯ is equal to three divided by π‘₯ squared plus four. And we call this a simple differential equation since we’ve written it in the form d𝑦 by dπ‘₯ is equal to some function of π‘₯. Now, to solve simple differential equations, we just need to integrate both sides of our equation with respect to π‘₯.

Remember that integration and differentiation are opposite processes. So the integral of d𝑦 by dπ‘₯ with respect to π‘₯ is just equal to 𝑦. So this will give us our solution 𝑦 up to our constant of integration 𝐢. We just need to evaluate the integral of three divided by π‘₯ squared plus four with respect to π‘₯. And to do this, we can notice that this integral is very similar to one of our integral rules involving the inverse trigonometric functions. For a constant π‘Ž not equal to zero, the integral of π‘Ž divided by π‘₯ squared plus π‘Ž squared with respect to π‘₯ is equal to the inverse tan of π‘₯ divided by π‘Ž plus our constant of integration 𝐢.

This is almost in this form already. For example, we can write four as two squared. We then just need our numerator to be two instead of three. And we can manipulate our expression so that this is true. We know that three over two multiplied by two is equal to three. So multiplying our entire integral by three over two and then changing our numerator to two does not change the value of our integral. However, it does allow us to evaluate this integral by using our integral rule. Our value of π‘Ž is equal to two. Using this, we get three over two times the inverse tan of π‘₯ over two plus the constant of integration we will call 𝐢 one.

Remember, we need to solve the differential equation. So we’re not looking for a general antiderivative. We’re looking for a particular solution. And to do this, we need to use the fact that the question tells us when π‘₯ is equal to two, 𝑦 is equal to zero. Substituting 𝑦 is equal to zero when π‘₯ is equal to two, we get zero is equal to three over two times the inverse tan of two divided by two plus 𝐢 one. Now, we can simplify this equation to find our value of 𝐢 one.

First, the inverse tan of two divided by two is the inverse tan of one, which we know is πœ‹ over four. Next we can just divide both sides of this equation by three over two. Finally, we just subtract πœ‹ by four from both sides of this equation. We get that 𝐢 one is equal to negative πœ‹ by four. All we need to do now is to use the value of 𝐢 one is equal to negative πœ‹ by four in our general solution.

Doing this, we get 𝑦 is equal to three over two times the inverse tan of π‘₯ over two minus πœ‹ by four. And we could leave our answer like this. However, we’ll distribute three over two over our parentheses. This gives us three over two times the inverse tan of π‘₯ over two minus three πœ‹ by eight. And this is our final answer. Therefore, we’ve shown the differential equation d𝑦 by dπ‘₯ times π‘₯ squared plus four is equal to three where 𝑦 of two is equal to zero has the solution 𝑦 is equal to three over two times the inverse tan of π‘₯ over two minus three πœ‹ by eight.

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