# Video: Solving a Differential Equation

Find the solution for the following differential equation for 𝑦(0) = 2 : 1 − 𝑒^(−𝑥) (d𝑦/d𝑥) = 0, 𝑦(0) = 2.

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### Video Transcript

Find the solution for the following differential equation for 𝑦 of zero is equal to two. One minus 𝑒 to the power of negative 𝑥 times d𝑦 by d𝑥 is equal to zero, where 𝑦 of zero is equal to two.

The question gives us a differential equation and an initial condition, and it wants us to find the solution to this differential equation. It’s worth noting we call this an initial condition because it’s when our variable 𝑥 is equal to zero. First, we need to know what it means to find a solution of a differential equation. We need to find 𝑦 is equal to some function 𝑓 of 𝑥 so that this satisfies our differential equation and our initial condition. 𝑦 equals 𝑓 of 𝑥 satisfying our differential equation means that one minus 𝑒 to the power of negative 𝑥 times 𝑓 prime of 𝑥 is equal to zero. And 𝑦 being equal to 𝑓 of 𝑥 satisfying our initial condition means that 𝑓 of zero must be equal to two.

So how are we going to find our function 𝑓 of 𝑥? Well, we can rearrange our differential equation to make 𝑓 prime of 𝑥 the subject. First, we add 𝑒 to the power of negative 𝑥 times 𝑓 prime of 𝑥 to both sides of the equation. We get one is equal to 𝑒 to the power of negative 𝑥 times 𝑓 prime of 𝑥. Next, we want to divide both sides of our equation through by 𝑒 to the power of negative 𝑥. And we know this is the same as just multiplying both sides of our equation by 𝑒 to the power of 𝑥. This gives us that 𝑓 prime of 𝑥 is equal to 𝑒 to the power of 𝑥.

But this is an equation for our derivative function 𝑓 prime of 𝑥. We want an expression for our function 𝑓 of 𝑥. And to do this, we need to remember that integration and differentiation are opposite processes. So we’ll integrate both sides of this equation with respect to 𝑥. So, because integration and differentiation are opposite processes, integrating 𝑓 prime of 𝑥 with respect to 𝑥 will give us 𝑓 of 𝑥. And it’s worth reiterating this will only be up to our constant of integration.

Now, let’s evaluate our other integral, the integral of 𝑒 to the power of 𝑥 with respect to 𝑥. Well, we know that this is just equal to 𝑒 to the power of 𝑥 plus 𝐶. So we’ve now found an expression for 𝑓 of 𝑥 up to our constant of integration 𝐶. It’s equal to 𝑒 to the power of 𝑥 plus 𝐶. Now, to find our value of 𝐶, we’re going to use our initial condition, 𝑦 of zero is equal to two or, equivalently, 𝑓 evaluated at zero is equal to two. Substituting this information into our expression for 𝑓 of 𝑥. We get that two is equal to 𝑒 to the zeroth power plus 𝐶. And of course, 𝑒 to the zeroth power is equal to one. So we can rearrange this equation and find that 𝐶 is equal to one.

So we’ve shown that 𝐶 is equal to one. So we’ll use this in our expression for 𝑓 of 𝑥. And this gives us that 𝑓 of 𝑥 is 𝑒 to the power of 𝑥 plus one. And remember, 𝑦 is equal to 𝑓 of 𝑥. So this gives us the equation 𝑦 is equal to 𝑒 to the power of 𝑥 plus one. And this is our final answer, we’ve shown the solution to the differential equation one minus 𝑒 to the power of negative 𝑥 times d𝑦 by d𝑥 is equal to zero and 𝑦 at zero is equal to two is 𝑦 is equal to 𝑒 to the power of 𝑥 plus one.