Video: Solving a First-Order Separable Differential Equation given Its Initial Condition

Find the solution of the differential equation d𝑦/dπ‘₯ = π‘₯𝑒^(𝑦) that satisfies the initial condition 𝑦(0) = 0.

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Video Transcript

Find the solution of the differential equation d𝑦 by dπ‘₯ equals π‘₯ times 𝑒 to the power of 𝑦 that satisfies the initial condition 𝑦 of zero equals zero.

A separable differential equation is a first-order differential equation in which the expression for d𝑦 by dπ‘₯ can be factored as a function of π‘₯ times a function of 𝑦. In other words, it can be written in the form d𝑦 by dπ‘₯ equals 𝑔 of π‘₯ times 𝑓 of 𝑦. Now, it’s quite clear that our equation d𝑦 by dπ‘₯ equals π‘₯ times 𝑒 to the power of 𝑦 is of this form. So, how do we solve it?

Well, we’re going to write it in differential form so that all of our 𝑦s are on one side and all of our π‘₯s are on the other. We’re going to divide through by 𝑒 to the power of 𝑦. And we’re going to split up our d𝑦 by dπ‘₯. Now, remember, this isn’t a fraction. But in this process, we do treat it a little bit like one. So, we find that one over 𝑒 to the power of 𝑦 d𝑦 equals π‘₯ dπ‘₯. Now, in fact, we’re going to write one over 𝑒 to the power of 𝑦 as 𝑒 to the power of negative 𝑦 just to make the next step easier.

And then, our next step is to integrate both sides of our equation with respect to their defined variable. So, on the left, we have the indefinite integral of 𝑒 to the power of negative 𝑦 with respect to 𝑦. And on the right, we have the indefinite integral of π‘₯ with respect to π‘₯. Now, if we recall the indefinite integral of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ for some constant values of π‘Ž is one over π‘Ž times 𝑒 to the power of π‘Žπ‘₯ plus some constant of integration 𝑐. So, when we integrate 𝑒 to the power of negative 𝑦 with respect to 𝑦, we get one over negative one times 𝑒 to the power of negative 𝑦 plus a constant 𝐴. And, of course, that’s just negative 𝑒 to the power of negative 𝑦 plus 𝐴. The integral of π‘₯ is fairly straightforward. We think of it as π‘₯ to the power of one. We then increase the exponent by one to give us two and divide by that new value. So, we get π‘₯ squared over two plus some other constant of integration 𝐡.

Let’s combine our constants of integration by subtracting 𝐴 from both sides. On the left-hand side, we’re simply left with negative 𝑒 to the power of negative 𝑦. And then on the right, we have π‘₯ squared over two plus a new constant 𝐢. Let’s multiply through by negative one. And we get negative π‘₯ squared over two plus a different constant. Remember, it’s essentially negative 𝐢. But let’s call that plus 𝐷. And then, we’re gonna find the natural log of both sides of our equation. And the reason we do that is because the natural log of 𝑒 to the power of negative 𝑦 is just negative 𝑦.

And so, we’re almost there with a general solution to our differential equation. To make 𝑦 the subject, we are gonna multiply through by negative one one more time. And so, we find the general solution to our differential equation to be 𝑦 equals negative the natural log of negative π‘₯ squared over two plus a constant 𝐷.

Well, there is one bit of information we haven’t used yet. We can find the particular solution to our differential equation by using the initial condition that 𝑦 of zero equals zero. In other words, when π‘₯ is equal to zero, 𝑦 is equal to zero. We replace π‘₯ and 𝑦 with zero. And we get zero equals negative the natural log of negative zero squared over two plus 𝐷. That gives us zero equals the negative natural log of 𝐷. But of course, we can multiply through by negative one. And we find that zero is equal to the natural log of 𝐷.

Now, we can observe a solution here. But we’re going to look at the full method. We’re going to raise both sides of this equation as a power of 𝑒. That gives us 𝑒 to the power of zero equals 𝑒 to the power of the natural log of 𝐷. Well, anything to the power of zero is one. And we know that 𝑒 to the natural log of 𝐷 is simply 𝐷. So, we find 𝐷 to be equal to one. And then, we can replace 𝐷 with one in our general solution.

And we now have a particular solution to the differential equation. It’s 𝑦 equals negative the natural log of negative π‘₯ squared over two plus one. And you might see this alternatively written as negative the natural log of negative one-half π‘₯ squared plus one.

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