Question

solve (x'2-ay)dx=(ax-y'2)dy​

Answer 1

Answer:

The solution of $(x^{2} - ay)dx = (ax - y^{2} )dy$ is derived as $x^{3} + y^{3} = 3axy + C$

Step-by-step explanation:

We have

$(x^{2} - ay)dx = (ax - y^{2} )dy\\x^{2}dx - aydx = axdy - y^{2}dy\\x^{2}dx - aydx - axdy + y^{2}dy = 0\\x^{2}dx + y^{2}dy = axdx + aydy\\x^{2}dx + y^{2}dy = a(xdx + ydy)$

From $xdy +ydx = d(xy)$ we get

$x^{2}dx + y^{2}dy = a(d(xy))$

So, we take integral on both sides

∫$x^{2} dx$ + ∫$y^{2}dy$ = $a$∫$d(xy)$

We get

$\frac{x^{3} }{3} + \frac{y^{3} }{3} = a(xy) + C$

By simplifying, we get the final solution

$x^{3} + y^{3} = 3axy + C$

I hope this answer may help you