Solve the homogeneous equations in Exercises 5–10. First put t...

Question

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.

(x²+y²)dx + xy dy = 0

Accepted Answer

Hello. In this video we are going to be solving the differential equation by reducing it to a homogeneous equation of the form D Y D X equal to a function of y divided by x. We may leave our answer in an implicit form. So the equation given to us is the quantity X2 minus Y2DX plus 2 XYDY is equal to 0. Now, the first thing we can do is we can go ahead and introduce the differential DYDX by dividing the entire equation by DX. So if we were to divide the entire equation by DX, we are left with X2 minus Y2 plus 2 XY DYDX, and that is equal to 0. Now the next thing we want to do is you want to go ahead and isolate the the differential DYDX. We can start by subtracting both sides of this equation by x2 minus y2. That will leave us with 2 X Y D Y D X. Equal to x2 minus Y2. Then we can go ahead and divide both sides of this equation by 2 XY. That will leave us with DYDX equal to X2 minus Y2, all divided by 2 XY. Now here, we can go ahead and further simplify the equation by separating it separating it into a sum or difference of fractions. So let's go ahead and start by distributing this -1 into the group X2 minus Y2. That would leave us with DYDX equal to Y2 minus X2, and this will all be divided by 2 XY. And if we were to separate this into a difference of fractions, we are left with D Y D X equal to y2 divided by 2 xy. Minus x 2 divided by 2 x y Further simplifying will allow us to simplify the differential equation to be Dyd x equal toy divided by 2 x minus x divided by 2y. Now keep in mind that we want our equation to be of the form y divided by X, so we can further simplify by dividing by the reciprocal of x divided by 2 y to the second fraction. That will allow us to rewrite the differential as D Y D X equal toy divided by 2 x minus. 1 divided by 2y divided by X. Now here what we can go ahead and do is we can further simplify the differential into a homogeneous equation by setting a substitution. Let's go ahead and allow you to equal toy divided by X. If we solve for y, we will also get y is equal to u multiplied by x, and if we were to take the derivative of this equation, we will get D Y D X is equal to u plus X D U D X. So we have a substitution for our differential DYDX, we have a substitution for any Y term, and we have a substitution for Y divided by X. So if we were to apply these substitutions, we can further rewrite the given differential as U plus X D U D X equal to U divided by 2. -1 divided by 2 u Now, again, what we want to go ahead and do is we want to solve for the differential symbol. So we will go ahead and subtract you to both sides of this equation. That will leave us with X D U D X equal to U divided by 2 minus 1 divided by 2 U minus U. Now, if we were to combine like terms together, we can further simplify the differential to be X D U D X equal to U divided by 2 minus 1 divided by 2 U. Combining the terms on the right hand side together will allow us to rewrite the equation to be X D U D X equal. To U2 + 1 divided by 2U. Now, what we want to go ahead and do is further simplify this differential equation by using separation of variables. So we are going to move all of the U terms to the left hand side of the equation, and move all of the X terms to the right hand side of the equation. So by moving the respective terms, we will end up getting. To you Divided by u2 + 1 dU equal to -1 divided by x dx. Now, in order to remove the differential symbols, we will need to integrate both sides of this equation. So the left hand side, I'm sorry, the right hand side is going to simplify and integrate to be negative natural logarithm of X + C. But on the left hand side, the left hand side is going to integrate to be the natural logarithm of U2 + 1. So by combining both parts together, we will end up with the natural logarithm. Of u2 + 1 and that will equal to the negative natural logarithm of x + c. Now what we can go ahead and do is use our natural logarithm properties to combine both of the natural logarithms together. So we will go ahead and add the natural logarithm of x to the left hand side. That will leave us with LN U 2 + 1 plus the natural logarithm of x, and this will equal to our generic constant c. By using the sum property, we can take the product of both of the arguments of the natural logarithm, and we will have the natural logarithm of X multiplied by U2 + 1, and this will equal to the generic constant C. Now remember, or what we can go ahead and do at this point is we can remove the natural logarithm by exponentiating both sides of this equation. Now, on the left hand side, that will remove our natural logarithm. On the right hand side, the exponential value of a constant is still going to be a constant. So this right hand side will still be a generic constant C. So we will be left with X multiplied by U2 + 1 equal to C. Now, what we will go ahead and do is apply our original substitution for you, and keep in mind that earlier we stated that U is equal to Y divided by X. So if we were to reapply our original substitution, we will be left with X multiplied by Y2 divided by X2 + 1, and that will equal to C. And if we were to combine the like terms in the group together, that will give us x multiplied by y2 + x2. All divided by x 2 equal to our generic constant c. And simplifying together. Will leave us with y2 + x2 divided by x equal to c. And if we were to just multiply both sides of this equation to remove the denominator, our homogeneous equation is going to be Y2 + X2 equal to CX. This can also be rewritten as x2 + y2 equal to c x, and this is going to be the simplest form of the homogeneous equation. And with that being said, the solution to this problem is going to be option C. 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 homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.(x²+y²)dx + xy dy = 0

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Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.

(x²+y²)dx + xy dy = 0