how to integrate:
y' = 2xy/(x^2-y^2)
first of all, the first y doesn't need a 1 exponent, so when you fix that error, you can ask that algebraic equation of a question.!.
I will assume that by y' you mean dy/dx
so it looks like your derivative is the result of an implicit derivative.
so le't work it backwards
y'(x^2 - y^2) = 2xy
y'x^2 - y'y^2 - 2xy = 0
looks like it could have been
(x^2)(y) - (1/3)y^3 = 0
let's differentiate:
x^2(y') + y(2x) - y^2y' = 0
sure enough!! it works
I must admit there was no real method to what I did, just some observation of patterns and lucky guessing.
To integrate the given differential equation:
y' = 2xy / (x^2 - y^2)
We can employ separation of variables. The basic idea is to rearrange the equation in a way that all the x terms are on one side and all the y terms are on the other side.
1. Begin by multiplying both sides of the equation by (x^2 - y^2) to eliminate the denominator:
(x^2 - y^2)dy = 2xy dx
2. Next, we split the equation into two separate fractions:
x^2 dy - y^2 dy = 2xy dx
3. Now we integrate both sides of the equation:
∫ (x^2 - y^2) dy = ∫ 2xy dx
To integrate the left-hand side, we can use the formula for the integral of a sum of functions, which is the same as integrating each term separately:
∫ x^2 dy - ∫ y^2 dy = ∫ 2xy dx
∫ x^2 dy = ∫ y^2 dy + ∫ 2xy dx
4. Evaluate the integrals individually:
∫ x^2 dy = y^3/3 + C1 (where C1 is the constant of integration)
∫ y^2 dy = y^3/3 + C2 (where C2 is another constant of integration)
∫ 2xy dx = x^2y + C3 (where C3 is yet another constant of integration)
5. Substitute the respective integral results back into the equation:
y^3/3 + C1 - y^3/3 - C2 = x^2y + C3
The constant term C1 and C2 cancel out:
0 = x^2y + C3
6. Finally, rearrange the equation to solve for y:
y = -C3 / x^2 (where -C3 is another constant of integration)
Therefore, the solution to the given differential equation is:
y = -C3 / x^2.