There is a process to find the general solution of X (1 + y Square) dx-y (1 + x square) dy = 0!

There is a process to find the general solution of X (1 + y Square) dx-y (1 + x square) dy = 0!

(1 + x square) dxx-y (1 + x square) dxx-y (1 + x square) DX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + y Square) DXX (1 + x square) DXX (1 + x square) DXY (1 + y (1 + y \\\\\\\\\\\\\\\\\\\\\\\\\\\\\= ln (...)

The general solution of dy / DX = x / y + Y / X

Let Y / x = u, y = Xu, y '= u + Xu' be substituted to obtain:
The general solution is: 1 / 2U ^ 2 = ln | x | + C

If the integral (2x + ax-y + 5) - (the square of 2bx-3x + 5y-1) is simplified without the square term and X term of X, then a = what, B = what
sorry. If the integral (2x's square + ax-y + 5) - (2bx's square-3x + 5y-1) is reduced without the square term and X term of X, then a = what and B = what

(2x²+ax-y+5)-(2bx²-3x+5y-1)=2x²+ax-y+5-2bx²+3x-5y+1
=(2-2b)x²+(a+3)x-6y+6
Square term without X and X term
So 2-2b = 0
a+3=0
So a = - 3, B = 1

Simplify "2x square + ax-y + 6-2bx square + 3x-5y-1"

2x²+ax-y+6-2bx²+3x-5y-1
= (2-2b)x²+(a+3)x-6y+5