### Video Transcript

Solve the differential equation π¦ prime plus π₯ times π to the power of π¦ is equal zero.

Weβre given a differential equation, and weβre asked to find the solution to this differential equation. The first thing we should do is try and write our differential equation in a form which we know how to solve. So to start, weβll write π¦ prime as dπ¦ by dπ₯. So weβve rewritten our differential equation as dπ¦ by dπ₯ plus π₯ times π to the power of π¦ is equal to zero. The next thing we could try and do is write our differential equation in the form dπ¦ by dπ₯ is equal to some function of π₯. Then we could just integrate both sides of the equation with respect to π₯.

However, we have a problem. We have π to the power of π¦ in our differential equation. And we canβt remove this. So weβll always going to have this function of π¦ in our differential equation. We can see that itβs multiplied by π₯. This should remind us of separable differential equations. Remember, we call a differential equation separable if it can be written in the form dπ¦ by dπ₯ is equal to some function π of π¦ times some function π of π₯. In our case, we can do this by subtracting π₯ times π to the power of π¦ from both sides of our differential equation. This gives us our function π of π¦ is π to the power of π¦ and our function π of π₯ is negative π₯. And the reason we call these separable differential equations is we can try and solve these by separating our variables.

To separate our variables, weβll start by dividing both sides of the equation through by π to the power of π¦. Using our laws of exponents, weβll write this as π to the power of negative π¦ times dπ¦ by dπ₯ is equal to negative π₯. Now, remember, dπ¦ by dπ₯ is not a fraction. However, when weβre solving separable differential equations, we can treat it a little bit like a fraction. This gives us the equivalent statement π to the power of negative π¦ dπ¦ is equal to negative π₯ dπ₯. Then all we need to do is integrate both sides of our equation. We get the integral of π to the power of negative π¦ with respect to π¦ is equal to the integral of negative π₯ with respect to π₯. We can now evaluate both of these integrals.

Letβs start with the integral of π to the power of negative π¦ with respect to π¦. We recall, for any constant π not equal to zero, the integral of π to the power of ππ¦ with respect to π¦ is equal to one over π π to the power of ππ¦ plus the constant of integration πΆ. We just divide by the coefficient of π¦ in our exponent. In our case, the coefficient of π¦ is negative one. So we want to divide our integrand by negative one. This gives us negative π to the power of negative π¦. And we could add a constant of integration here. However, remember, weβre also going to integrate negative π₯ with respect to π₯. So weβll just combine these two constants of integration together.

Now, we want to evaluate the integral of negative π₯ with respect to π₯. Remember, negative π₯ is actually negative π₯ to the first power. So we can integrate this by using the power rule for integration, which tells us, for any constant π not equal to negative one, the integral of π₯ to the πth power with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus the constant of integration πΆ.

We want to add one to our exponent of π₯ and then divide by this new exponent. In our case, the exponent of π₯ is one. So we add one to this to get two and then divide by two. This gives us negative π₯ squared divided by two plus the constant of integration weβll call πΆ one.

And this is a general solution to our differential equation. However, this is an implicit solution because itβs not of the form π¦ is equal to some function of π₯. So if possible, we should try to write this in the form π¦ is equal to some function of π₯. So letβs try and do this. Weβll start by multiplying both sides of our equation through by negative one. And remember, πΆ one is a constant. So we can write negative πΆ one as a new constant weβll call πΆ. So we now have π to the power of negative π¦ is equal to π₯ squared over two plus πΆ.

Remember, we want to rewrite this equation so π¦ is the subject. So weβll need to take the natural logarithm of both sides of our equation. Doing this, we get the natural logarithm of π to the power of negative π¦ is equal to the natural logarithm of π₯ squared over two plus πΆ. Now, remember, the natural logarithm function and the exponential function are inverses. So the natural logarithm of π to the power of negative π¦ would just be equal to negative π¦. So we now have negative π¦ is equal to the natural logarithm of π₯ squared over two plus πΆ. And the last thing we need to do is multiply both sides of this equation through by negative one.

And this means weβve written our general solution in the form π¦ is equal to some function of π₯. Therefore, given the differential equation, π¦ prime plus π₯ times π to the power of π¦ is equal to zero, we were able to find the solution π¦ is equal to negative one times the natural logarithm of π₯ squared over two plus πΆ.