Solve the differential equation in Exercises 9–22.
19. y...

Question

Solve the differential equation in Exercises 9–22.

19. y²(dy/dx) = 3x²y³ - 6x²

Accepted Answer

Welcome back everyone. In this problem we want to solve the differential equation that says Y2 multiplied by the derivative of y with respect to x equals 2xy cubed minus 4 x. Now for us to solve this equation, let's first try to separate our variables to get all the y variables on one side and to get the x variables on one side of our equation. No, given our equation, if we factor 2 x out of the right side, then we'll have y2 multiplied by DYD X. Being equal to 2 x multiplied by y cubed minus 2. And if we divide both sides by y cubed minus 2 and multiply by dx, then Y2 divided by y cubed minus 2 Dy will be equal to 2 xDX. Now we can integrate both sides to help us solve for why. How can we do that? Well, let's take our left side, OK, so. Integrating. OK, integrating y2 divided by y cubed minus 2 with respect toy, we can use substitution. So to do it, let's let you be equal to y cubed minus 2. OK. Then that means the derivative of u with respect toy would be equal to 3y2. So duu would be equal to 3y2 Dy. And thus that means then Y 2 Dy would be equal to 1/3 of DU. So our integral then can be rewritten in terms of you as the integral of 1 divided by 3 you with respect to you. That could be 1/3 of the integral of 1 divided by u with respect to u, which would be equal to 1/3 of the natural log of the absolute value of u, which we already know is y cubed minus 2. So this would be our result on the left hand side of our equation. What about on our right side? Well, on the right side, the integral of 2 x with respect to x would be equal to x 2 plus some constant c1. So now that we have both sides, equating the sides gives us 1/3 of the natural log of y cubed minus 2. To be equal to x 2 plus some constant c1, now we can use this to solve explicitly for y. Now if we multiply by 3, then we get the natural log of y cubed minus 2 to be equal to 3 x2 plus 3c1. And if we exponentiate both sides. Then we get the absolute value of y cubed minus 2. To be equal to e to the power of 3 x2 multiplied by e. To the power of 3C1, OK, because remember that here if both of these were in the power 4e, then we can rewrite it as the base is being multiplied. OK, so in this case if we simplify, then that means y cubed minus 2 will be equal to plus or minus e to the power of 3c1 multiplied by e to the power of 3 x 2. And now if we let C be equal to plus R minus E to the power of 3C1, then that means Y cubed minus 2 could be rewritten as CE to the power of 3 x 2, OK. Where c equals plus r minus e to the 3C1. Well, in that case, OK, then that means Y cubed. Would be equal to c e to the power of 3 x 2 OK plus 2 and now if we take the cube root off both sides. Then that meansy would be equal to the cube root of c e to the power of 3 x2 + 2. Thus this would be the solution to our differential equation. Thanks a lot for watching everyone. I hope this video helped.

Solve the differential equation in Exercises 9–22.19. y²(dy/dx) = 3x²y³ - 6x²

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Solve the differential equation in Exercises 9–22.
19. y²(dy/dx) = 3x²y³ - 6x²