Video: Solving Differential Equations

Find the solution for the following differential equation for 𝑦(0) = 1: (d𝑦/dπ‘₯) βˆ’ π‘₯ βˆ’ π‘₯Β² = 1.

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Video Transcript

Find the solution for the following differential equation for 𝑦 of zero equals one. d𝑦 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, 𝑦 is equal to one. Now, before we can look to solve this differential equation, we’re going to need to perform an intermediate step. We’re going to rearrange and make d𝑦 by dπ‘₯ the subject. Now, it’s not particularly difficult to achieve this. We add π‘₯ and π‘₯ squared to both sides of our differential equation to find that d𝑦 by dπ‘₯ equals π‘₯ squared plus π‘₯. Now, remember, we can find a general solution for the differential equation d𝑦 by dπ‘₯ equals π‘₯ squared plus π‘₯ by performing the reverse process to differentiation. We’re going to integrate with respect to π‘₯. So 𝑦 will be equal to the integral of d𝑦 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 we’re working with indefinite integrals, we know we need that constant of integration 𝑐. So 𝑦 is equal to π‘₯ cubed over three plus π‘₯ squared over two plus 𝑐. So we have the general solution to our differential equation. But we haven’t yet used the information that 𝑦 of zero equals one. We’re going to substitute π‘₯ equals zero and 𝑦 equals one into this general solution. When we do, we see that one equals zero cubed over three plus zero squared over two plus 𝑐. And this whole equation simplifies to one equals 𝑐. So we found that 𝑐 equals one. And the particular solution to our differential equation is 𝑦 equals π‘₯ cubed over three plus π‘₯ squared over two plus one.

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