In Exercises 23–28, solve the initial value problem.
y d...

Question

In Exercises 23–28, solve the initial value problem.

y dx + (3x - xy + 2)dy = 0, y(2) = -1, y < 0

Accepted Answer

Solve the initial value problem. XDY minus Y + XY + 1 DX equals 0, with Y 1 equals 0 and X greater than 0. Now, this is a separable variables problem, so let's separate our variables. We'll do this by moving DY to one side and DX to the other side. We can say XDY. Equals Y + X Y + 1. D X Now from here. We still need to group X and Y together and do some factoring. So, On the right side, we will factor out why. XDY equals Y multiplied by 1 + X + 1. DX. Now, let's separate our variables. We can divide both sides with the left by x. Giving us d y equals y multiplied by 1 + x + 1. Divided by X and DX. And then we divide by the numerator. To get DY Divided by y multiplied by 1 + x + 1. Equals DX divided by X. From here, we can take our antiderivative of both sides. Now, let's take the inside derivative of both. On the right side, we have the integral DX divided by X, which is just natural log, absolute value of X plus C. Now we need the integral of the left side, which we will use a substitution. We have the integral DY divided by Y multiplied by 1 + X + 1. Now for here, we'll let U equals Y multiplied by 1 + X + 1. And do you, since we want this in terms of why, we'll take the derivative of Y. This just turns into 1 divided by 1 + X. DU, we will say this is actually DY. Now, from here, we will use our substitution. We have the integral. Which in this case, is 1 divided by 1 + X or RDY. Multiplied by the integral 1 + XDY. Divided by 1 + x multiplied by y + 1. We can then solve this integral. We ended up getting 1 divided by 1 plus X. Multiplied by DU divided by U. Which is equals to 1 divided by 1 + X, natural log absolute value of U. Now we can substitute our you back in. We have 1 divided by 1 + x natural log of y multiplied by 1 + x + 1. Plus c Now, we can set our integrals equal to each other, since they were on both sides of the equal sign. We have 1 divided by 1 + X, natural log, absolute value Y, 1 + X + 1. Equals Natural log, absolute value X plus C. Also called the sea one. Now, we can solve for our particular solution by using our initial condition. We have y1 equals 0. Which means X equals 1, Y equals 0. So we have 1 divided by 1 plus one. Natural light 0 multiplied by 1 + 1 + 1. Equals natural log of 1 plus C1. Simplifying all of this gives us C1 equals 0. We can then substitute this back into our equation. We have 1 divided by 1 + X, natural log, absolute value Y, 1 + X + 1. Equals natural logx plus 0. Now from here, we just need to solve for Y. And we do know that X is greater than 0, so we can remove our Absolute values on our second natural luck. Now from here, let's take E on both sides and exponentiate. We ended up getting the absolute value. 1 + x multiplied by y + 1. Equals E to the 1 + X natural log X. We also multiplied the 1 + X over to the other side. We can then simplify even further. We get 1 + X, Y + 1. Equals And now, for our right son. We get plus or minus X-ray C1 plus X. By simplifying our e to the natural log of X. With our initial condition, we can only use the positive value, so we will just use the positive. Now we just need to isolate why. We can subtract 1 and divide by 1 + x to get y equals. X 1 + X minus 1. Divided by X + 1. This will be the solution to our initial value problem. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 23–28, solve the initial value problem.y dx + (3x - xy + 2)dy = 0, y(2) = -1, y < 0

Watch next

In Exercises 23–28, solve the initial value problem.
y dx + (3x - xy + 2)dy = 0, y(2) = -1, y < 0