Solve the Bernoulli equations in Exercises 29–32.

Question

x²y' + 2xy = y³

Accepted Answer

Hello. In this video, we are going to be solving for the following Bernoulli equation. Now, before we approach this problem, let's go ahead and review the Bernoulli equation generally. A general Bernoulli equation is of the form Y plus a function of XP X multiplied by Y equal to a function of XQ X multiplied by Y raised the power of n. Now as we see in our Bernoulli equation, Y has a coefficient of 1. In our problem, we have a coefficient of X, which we need to simplify. So to start this problem, we are going to divide the entirety of our Bernoulli equation by X. That will allow us to rewrite the equation as Y plus 1 divided by X multiplied by Y, and that is equal to x multiplied by Y2. Now here, in order to simplify the Bernoulli equation, we want to linearize the Bernoulli equation. The Bernoulli equation has an identity V equal to y raise the power of 1 minus n, and this is going to help us linearize the equation. Now n is defined as the exponential value of y on the right hand side of the equation, which in this case is 2. So we are going to have n is equal to 2. By plugging this in, we will get v is equal to y raise the power of 1 minus 2, and that will give us v is equal to y to the power of -1. Now here, we need to go ahead and find a substitution for Y in terms of V. So if we take the derivative of both sides of this equation, we get V is equal toY raised the power of -2 Y. So If we take this substitution here and apply this to our Bernoulli equation, we can rewrite the equation to be wise to the -2 multiplied by V plus 1 divided by X. Now here we also need a substitution for Y. Notice that we have v is equal to y to the power of -1. This is equivalently written as v is equal to 1 divided by y, and by solving for y we get y is equal to v to the negative first power. So, this will be replaced with V to -1. And that will equal to x multiplied by y 2. Now what we want to do here is make sure that our differential symbol v. Also has a coefficient of 1, so we will need to divide the entirety of this equation by y to the -2. That is going to leave us with V minus V 1st divided by X multiplied by Y2 equal to X or just X on the right hand side. We need to simplify this middle term, and we need to simplify it by coming up with a solution or a simplification of y squared. So if we take a closer look at our second term, we have negative V to the 1st divided by X multiplied by Y2. Now this y square in the denominator is equivalent toy to the power of -2. If we take our y substitution earlier and raise both sides to the negative 2nd power, we get y to the -2 is equal to v2. So, what this means is that we can replace this denominator with V2. That can allow us to rewrite the second term to be negative V to the negative1 divided by X multiplied by V 2, and this is going to simplify to be 1 divided by X multiplied by V. So let's go ahead and replace that in our equation. So we have V minus 1 divided by X multiplied by V equal to X. Now what we've done here is we've simplified the Bernoulli equation into a first-order linear differential equation. So we have a function of X, P of X, which is going to be -1 divided by X, and the right-hand side is going to be Q of X. So in order to simplify, we need to go ahead and identify our integrating factor. The integrating factor is defined as m is equal to ease the power of p x. That is going to simplify to be e raised the power of the integral of -1 divided by x which will give us e raised the power of negative natural logarithm of x which will further simplify to be x raised to the power of -1. So, what we're going to do here is we're going to take our integrating factor and multiply it to the entirety. Of our differential equation That is going to allow us to rewrite the differential equation to be 1 divided by X multiplied by V minus 1 divided by X2 multiplied by V. Equal to -1. Now here, this is a little tricky to see, but notice that the left hand side of this equation is equivalent to a derivative. This is equivalent to the derivative with respect to x of the product of 1 divided by x multiplied by V. So we can rewrite the left hand side of our equation as a derivative, that will be the derivative with respect to X of 1 divided by X multiplied by V equal to -1. Now we want to go ahead and remove the differential symbol, and in order to do that, we will have to integrate both sides of this equation. That would leave us with 1 divided by X multiplied by V, and that is equal to negative X plus the generic constant C. Now we want to isolate V on the left hand side, and in order to do that, we will multiply the entirety of the equation by x. That will give us V is equal. To g x 2 plus c multiplied by x, and if we reorder the terms, we give v is equal to c multiplied by x minus x 2. Now, the last thing we want to do is you want to resubstitute V back into terms of Y. Now, earlier, we had a substitution that V is equal to -y, and this is equivalent to saying that V is equal to 1 divided by Y. So if we take the substitution and plug this back into the general equation or general solution, we get 1 divided by Y is equal to CX minus X2. And finally, solving for y gives us Y is equal to 1 divided by CX minus X2, and this is going to be the general solution to the Bernoulli equation. And the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Solve the Bernoulli equations in Exercises 29–32.x²y' + 2xy = y³

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Solve the Bernoulli equations in Exercises 29–32.

x²y' + 2xy = y³