### Video Transcript

Find the solution for the following
differential equation for π¦ of zero equals one. dπ¦ by dπ₯ minus π₯ minus π₯
squared equals zero.

In this question, we have a
differential equation and an initial value. That is, when π₯ is equal to zero,
π¦ is equal to one. Now, before we can look to solve
this differential equation, weβre going to need to perform an intermediate step. Weβre going to rearrange and make
dπ¦ by dπ₯ the subject. Now, itβs not particularly
difficult to achieve this. We add π₯ and π₯ squared to both
sides of our differential equation to find that dπ¦ by dπ₯ equals π₯ squared plus
π₯. Now, remember, we can find a
general solution for the differential equation dπ¦ by dπ₯ equals π₯ squared plus π₯
by performing the reverse process to differentiation. Weβre going to integrate with
respect to π₯. So π¦ will be equal to the integral
of dπ¦ by dπ₯ with respect to π₯ or the integral of π₯ squared plus π₯ with respect
to π₯.

Remember, to integrate a polynomial
term, we add one to the exponent and then divide by that new value. So the integral of π₯ squared is π₯
cubed over three. We then see that the integral of π₯
is π₯ squared over two. And since weβre working with
indefinite integrals, we know we need that constant of integration π. So π¦ is equal to π₯ cubed over
three plus π₯ squared over two plus π. So we have the general solution to
our differential equation. But we havenβt yet used the
information that π¦ of zero equals one. Weβre going to substitute π₯ equals
zero and π¦ equals one into this general solution. When we do, we see that one equals
zero cubed over three plus zero squared over two plus π. And this whole equation simplifies
to one equals π. So we found that π equals one. And the particular solution to our
differential equation is π¦ equals π₯ cubed over three plus π₯ squared over two plus
one.