A calculus problem f (x + y, X-Y) = XY, then what are the partial derivatives of F for X and F for y?

A calculus problem f (x + y, X-Y) = XY, then what are the partial derivatives of F for X and F for y?

Let u = x + y, v = X-Y, then xy = (1 / 4) (U + V) (U-V) = (u ^ 2-V ^ 2) / 4
df/dx=df/du*du/dx+df/dv*dv/dx=0.5u-0.5v=y
df/dy=0.5u+0.5v=x

1. Y '= x (1-y ^ 2) ^ (1 / 2). Find y 2. (x ^ 2-9) y' + xy = 0, y (5) = 1. Find y

Solve the differential equation: 1. Y '= x √ (1-y & # 178;); find y 2. (x ^ 2-9) y' + xy = 0, y (5) = 1. Find y
1.dy/dx=x√(1-y²)
Dy / √ (1-y & # 178;) = xdx is obtained by separating variables, arcsiny = (1 / 2) x & # 178; + C is obtained by integrating both sides separately, that is, y = sin [(1 / 2) x & # 178; + C] is
Its general solution
2.(x²-9)y'+xy=0,y(5)=1；
(X & # 178; - 9) (dy / DX) = - XY, separate variables to get: (1 / y) dy = - xdx / (X & # 178; - 9); that is, (1 / y) dy = - (1 / 2) d (X & # 178; - 9) / (X & # 178; - 9)
Integral, LNY = - (1 / 2) ln (X & # 178; - 9) + LNC, so the general solution is y = C / √ (X & # 178; - 9); substitute y (5) = 1 to get 1 = C / √ (25-9)
So C = 4, and the special solution is y = 4 / √ (X & # 178; - 9)