# Video: Solving a Differential Equation

Find the solution for the following differential equation for π¦(0) = 2 : 1 β π^(βπ₯) (dπ¦/dπ₯) = 0, π¦(0) = 2.

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### Video Transcript

Find the solution for the following differential equation for π¦ of zero is equal to two. One minus π to the power of negative π₯ times dπ¦ by dπ₯ is equal to zero, where π¦ 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. Itβs worth noting we call this an initial condition because itβs when our variable π₯ is equal to zero. First, we need to know what it means to find a solution of a differential equation. We need to find π¦ is equal to some function π of π₯ so that this satisfies our differential equation and our initial condition. π¦ equals π of π₯ satisfying our differential equation means that one minus π to the power of negative π₯ times π prime of π₯ is equal to zero. And π¦ being equal to π of π₯ satisfying our initial condition means that π of zero must be equal to two.

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

But this is an equation for our derivative function π prime of π₯. We want an expression for our function π of π₯. And to do this, we need to remember that integration and differentiation are opposite processes. So weβll integrate both sides of this equation with respect to π₯. So, because integration and differentiation are opposite processes, integrating π prime of π₯ with respect to π₯ will give us π of π₯. And itβs worth reiterating this will only be up to our constant of integration.

Now, letβs evaluate our other integral, the integral of π to the power of π₯ with respect to π₯. Well, we know that this is just equal to π to the power of π₯ plus πΆ. So weβve now found an expression for π of π₯ up to our constant of integration πΆ. Itβs equal to π to the power of π₯ plus πΆ. Now, to find our value of πΆ, weβre going to use our initial condition, π¦ of zero is equal to two or, equivalently, π evaluated at zero is equal to two. Substituting this information into our expression for π of π₯. We get that two is equal to π to the zeroth power plus πΆ. And of course, π to the zeroth power is equal to one. So we can rearrange this equation and find that πΆ is equal to one.

So weβve shown that πΆ is equal to one. So weβll use this in our expression for π of π₯. And this gives us that π of π₯ is π to the power of π₯ plus one. And remember, π¦ is equal to π of π₯. So this gives us the equation π¦ is equal to π to the power of π₯ plus one. And this is our final answer, weβve shown the solution to the differential equation one minus π to the power of negative π₯ times dπ¦ by dπ₯ is equal to zero and π¦ at zero is equal to two is π¦ is equal to π to the power of π₯ plus one.