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Question Video: Solving First-Order First-Degree Linear Differential Equations Mathematics

Solve the differential equation π‘₯(d𝑦/dπ‘₯) = 𝑦 + π‘₯Β² sin π‘₯ subject to the condition 𝑦(πœ‹) = 0.

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Video Transcript

Solve the differential equation π‘₯ d𝑦 by dπ‘₯ equals 𝑦 plus π‘₯ squared sin π‘₯ subject to the condition 𝑦 of πœ‹ equals zero. Is it (a) 𝑦 equals π‘₯ cos π‘₯ plus π‘₯? (b) 𝑦 equals π‘₯ cos π‘₯ minus π‘₯. (c) 𝑦 equals negative π‘₯ cos π‘₯ plus π‘₯. (d) 𝑦 equals π‘₯ cos π‘₯. Or (e) 𝑦 equals negative π‘₯ cos π‘₯ minus π‘₯.

A first-order linear differential equation in its most general form looks like this: d𝑦 by dπ‘₯ plus some function 𝑓 of π‘₯ times 𝑦 equals some function 𝑔 of π‘₯. In most cases for 𝑓 of π‘₯ and 𝑔 of π‘₯, it will not be possible to separate the variables of this differential equation. So we need to use an integrating factor, which is given by 𝐼 of π‘₯ equals 𝑒 to the integral 𝑓 of π‘₯ dπ‘₯. And we ignore the constant of integration. We then multiply both sides of the differential equation by this integrating factor. What this achieves is that we now have an exact differential equation, which is to say that we have a derivative of only one term on the left-hand side.

More specifically, we have d by dπ‘₯ of the integrating factor times 𝑦. This then allows us to separate the variables and integrate with respect to π‘₯ on both sides. This gives us a standard formula for the solution to a general first-order differential equation, which may usually be quoted without proof, 𝐼𝑦 equals integral 𝐼𝑔 of π‘₯ dπ‘₯.

So let’s apply this to the given differential equation. We first need to rearrange the equation to get it in the standard form. That is, d𝑦 dπ‘₯ plus some function of π‘₯ times 𝑦 equals another function of π‘₯. Rearranging into this form, we get d𝑦 by dπ‘₯ minus 𝑦 over π‘₯ equals π‘₯ sin π‘₯. So in this case, our 𝑓 of π‘₯ is negative one over π‘₯ and our 𝑔 of π‘₯ is just π‘₯ sin π‘₯. So our integrating factor is given by 𝑒 to the integral 𝑓 of π‘₯ dπ‘₯, which in this case is 𝑒 to the integral negative one over π‘₯ dπ‘₯. This integrates to 𝑒 to the negative natural log of π‘₯. And we ignore the constant of integration.

Using standard indices laws, this is equal to 𝑒 to the natural log of π‘₯ to the negative one. 𝑒 to the natural log of π‘₯ is simply 𝑒 raised to the power of 𝑒 that makes π‘₯. So this is simply π‘₯. So our integrating factor 𝐼 of π‘₯ is one over π‘₯. Now plugging into the standard formula 𝐼𝑦 equals integral 𝐼𝑔 of π‘₯ dπ‘₯. We have one over π‘₯ 𝑦 equals the integral of one over π‘₯ times π‘₯ sin π‘₯ dπ‘₯. The right-hand side simplifies to the integral of sin π‘₯ dπ‘₯, which integrates to negative cos π‘₯. And this time, we do not ignore the constant of integration. Multiplying both sides by π‘₯, we get 𝑦 equals negative π‘₯ cos π‘₯ plus 𝐢π‘₯.

This is the general solution to the differential equation. But we’re not finished yet since this is subject to a condition 𝑦 of πœ‹ equals zero. Plugging in these values of π‘₯ and 𝑦, we get zero equals negative πœ‹ cos πœ‹ plus πΆπœ‹. Rearranging and simplifying, we get 𝐢 is equal to negative one. Substituting in this value of 𝐢 into the general solution, we get the particular solution 𝑦 equals negative π‘₯ cos π‘₯ minus π‘₯. Comparing this with our possible answers, we can see that this matches with (e). 𝑦 equals negative π‘₯ cos π‘₯ minus π‘₯.

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