Solve the Bernoulli equations in Exercises 29–32.

Question

y' - y = xy²

Accepted Answer

Hello. In this video, we are going to be finding the general solution to the Bernoulli equation Y minus 3Y equal to XY2. In order to approach this problem, let's review the general form of the Bernoulli equation. Now the general form is Y plus a function of X P X multiplied by Y equal to a function of Q X, which is QX multiplied by Y raised to the power of N. Now, in order to simplify the Bernoulli equation, the first thing we can do is we can try to linearize the Bernoulli equation. In order to linearize the Bernoulli equation, we want to come up with a substitution for Y. So, we will use the identity V is equal to y raised the power of 1 minus n. And again, the value of n comes from the exponential value of y on the right hand side of the equation. Now, if we take a look at our equation, the exponential value of Y of the right hand side is 2. So we're going to plug in the value and is equal to 2. That will give us v is equal to y to the power of 1 minus 2, which will simplify to be v is equal toy the power of -1. Now, if we were to take this statement and take the derivative of both sides of this equation, we will get V is equal to Y 2 multiplied by Y. And if we were to solve for Y, Y is going to equal to Y2 multiplied by V. So now what we're going to do is we're going to linearize the equation by applying the substitution for Y. That is going to allow us to rewrite the equation to be negative Y2 multiplied by V. Minus 3 multiplied by Y equal to x multiplied by Y 2. Now here we want to allow. V to have a coefficient of 1, so we will divide the entirety of the equation by negative y 2. That is going to leave us with V + 3 divided by Y equal to x. Now here, notice that we have 3 divided by Y. That is equivalent to 3 multiplied by y raised the power of -1, which we do have a solution for here. We can substitute that value in the denominator with just v. So we can further rewrite this Bernoulli equation as v plus 3 V equal to x. Now one thing to notice here is that we've simplified this equation to be a first order linear. Differential equation. And so in order to simplify the first order linear differently quenching, we need to we need to solve for an integrating factor. So, the integrating factor is going to be m is equal toe raises the power of the integral of p x, and P x is going to be the function of x in front of our v term in this case, so p of x is going to equal to the constant of 3. So our integrating factor is going to be equivalent to E raised to the power of the integral of 3, and that will integrate to bee raised to the power of 3 X. Now in order to simplify this equation, we are going to multiply the entirety of our equation by our integrating factor. That is going to leave us with E raised to the power of 3 x multiplied by V plus 3V multiplied by E raised to the power of 3 X, and that is equal to negative X multiplied by E raised to the power of 3 X. Now this is a little tricky to see at first, but the left hand side of the equation is equivalent to a derivative. Specifically, it is equivalent to the derivative of the product of E 3 x multiplied by V. So we can rewrite the left-hand side of this equation as a derivative. That will leave us with DDX multiplied by E raise the power of 3 XV, and that is equal to X multiplied by E raise the power of 3 X. Now, the next thing we want to do is we want to go ahead and remove the differential symbol, but in order to do this, we will have to integrate both sides of this equation. Now integrating the left hand side will eliminate the differential symbol, but we're going to need to integrate by parts in order to simplify the right hand side. So, we will go ahead and use integration by parts. We have the integral of x multiplied by e raised the power of 3 x. We will allow you to equal to x and dV equal to e raised to the power of 3 x. Differentiating the U statement gives us DU equal to negative DX, and integrating the DV statement gives us V is equal to 1/3 e to the power of 3 X. So, putting this into the identity for integration by parts gives us U multiplied by V, which is 1/3 X, E raised to the power of 3X minus the integral of V multiplied by DU, which is 1/3, E to the 3 XDX. And this will further simplify to be -1/3 XE to the power of 3x plus 1/9e the power of 3x plus the generic constant C. So, by applying the simplification to the right hand side, the overall differential equation is going to simplify to be e to the power of 3 x multiplied by V equal to 1/3 x e to the power of 3 x plus 1/9 e to the power of 3 x plus c. Now what we want to do is we want to make sure that ourv term has a coefficient of 1 on the left-hand side. In order to remove e to the power of 3 x, we will multiply the entirety of the equation by e to the power of -3 x. This will simplify the linear differential equation to be V is equal to -1/3 x + 1/9 + C multiplied by e to the power of -3 x. Now, what we want to do is we want to go ahead and backs substitute V back into terms of Y. Now earlier in the problem, we stated that V is equal to y to the power of -1. While here if we solve for y Or here what we're going to do is we're going to take V and replace it with Y to -1, but that is equivalent to V equal to 11 divided by Y. So, plugging this back into our differential equation gives us 1 divided by Y equal to -1/3 X + 1/9 + C multiplied by E to the power of -3X. And solving for y will give us the solution y is equal to 1 divided. By 1/3 x + 1/9 + c multiplied by e to the power of -3 x. This is going to be the general solution. To the given Bernoulli problem. And with that being said, 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.y' - y = xy²

Watch next

Solve the Bernoulli equations in Exercises 29–32.

y' - y = xy²