### Video Transcript

Find the solution for the following differential equation for π¦ of zero equals zero. dπ¦ equals π to the power of π₯ add π¦ dπ₯.

Weβre going to need to do some integration here. But remember, in order to do that, we need to have functions of π₯ and dπ₯ on one side and functions of π¦ and dπ¦ on the other side. But how can we separate the variables here when we have a function of π₯ and π¦? To do this, recall the product rule for exponents. π to the power π multiplied by π to the power π equals π to the power of π add π. And so, we can rewrite this differential equation as dπ¦ equals π to the power π₯ multiplied by π to the power π¦dπ₯.

And now, we can separate the variables. To do this, we divide through by π to the power π¦. So, now, we have functions of π¦ and dπ¦ on the left and functions of π₯ and dπ₯ on the right. So, now, we can integrate both sides. Letβs start by putting this one over π to the power π¦ in negative exponent form, which is π to the power of negative π¦. Letβs integrate the right-hand side first. This is the integral of π to the power π₯ with respect to π₯. To do this, recall an important general integral. The integral of π to the power ππ₯ with respect to π₯ equals one over ππ to the power ππ₯ add a constant of integration π.

So, the integral of π to the power π₯ is going to be π to the power of π₯ plus a constant of integration, which Iβll call π one. So, now, letβs integrate π to the power of negative π¦ with respect to π¦. We can think of this power negative π¦ as negative one π¦ if that helps. So, applying the same general integration rule as before, but this time weβve got π¦ and π is negative one, we get the π to the negative π¦ integrates to one over negative one π to the power of negative π¦ add a constant of integration, which this time we can call π two. One over negative one is just negative one. And we can combine our constants of integration π one and π two with just one constant of integration, which weβll call π.

Weβre not actually done with this question yet because we were given the condition that π¦ of zero equals zero. This means that when π₯ equals zero, π¦ equals zero. So, we substitute these values into our equation. And we remember that any non-zero number raised to the power of zero is one. So, π to the power zero is one. This gives us that negative one equals one add π. And so, π equals negative two. So, we substitute this into our equation. Finally, we can rearrange our equation to get our functions of π₯ and π¦ on the same side of the equation. This gives us the π to the power π₯ add π to the power of negative π¦ equals two.