Question Video: Solving Differential Equations | Nagwa Question Video: Solving Differential Equations | Nagwa

# Question Video: Solving Differential Equations

Find the solution for the following differential equation for π¦(0) = 1: (dπ¦/dπ₯) β π₯ β π₯Β² = 1.

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### 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, weβre going to need to perform an intermediate step. Weβre going to rearrange and make dπ¦ by dπ₯ the subject. Now, itβs 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. Weβre 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 weβre 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 havenβt yet used the information that π¦ of zero equals one. Weβre 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.

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