How to solve the equation axdy = xdx + YDY? How to solve the differential equation axdy = xdx + YDY? You don't need to calculate the specific value, just find out the relationship between X and y Give a specific process, thank you

How to solve the equation axdy = xdx + YDY? How to solve the differential equation axdy = xdx + YDY? You don't need to calculate the specific value, just find out the relationship between X and y Give a specific process, thank you

Axdy = xdx + ydyxdx = (ax-y) dy the relationship between X and y can be obtained by integrating both sides of Dy (but we should regard x on the right side of the equal sign as a function of Y)

The general solution of xdx + e ^ YDY = 0

e^ydy=-xdx
e^y=-X^2/2+C

The general solution of the differential equation dy / DX = Y2 / X is

dy/y^2 = dx/x
=>-1/y = lnx +C
=>lnx+1/y+C=0
=> y = -1/(lnx+C)

General solution of differential equation XDY YDX = 0 (detailed process) thank you!

Separate variable method
 

The general solution of the differential equation xdy-3ydx = 0 is?

xdy-3ydx=0
dy/y=3dx/x
ln(y)=3ln(x)+c
ln(y/(cx^3))=0
y=cx^3
C is an arbitrary constant

The general solution of YLN xdx + XLN YDY = 0

lnx/xdx=-lny/ydy
lnxd(lnx)=-lnyd(lny)
1/2(lnx)^2=-1/2(lny)^2+C

(-4/9)xy…… 3+（8/27）x…… 3 y…… Factoring For several times of a certain number (such as X 3 is the third power of X

(-4/9)xy³+（8/27）x³ y²=(-4/9)xy²(y-2/3x²)

When k = (), the second power of X - (the second power of 3kxy-3y) + one third of XY-8 does not contain XY term, is it 1 / 9 or - 1 / 9

The guarantee is: the second power of X - (the second power of 3kxy-3y) + one third of XY-8
The total term containing XY is - 3kxy + 1 / 3xy = 0, then - 3K + 1 / 3 = 0, then 3K = 1 / 3, k = 1 / 9

The octave of x plus the octave of Y, x + y = 4, xy = 4
Such as the title

X + y = xy = 4 find X & sup2; + Y & sup2;, x ^ 4 + y ^ 4, x ^ 8 + y ^ 8
x²+y²=(x+y)^2-2xy=8
x^8+y^8=(x^4+y^4)^2-2x^4y^4=512
Or directly solve x = y = 2

Who knows how to find the partial derivative of Z = (1 + XY) to the Y power
Thank you

Logarithmic derivation method
Take logarithm on both sides: lnz = y * ln (1 + XY)
The partial derivatives of two sides for Y: z'y / z = ln (1 + XY) + y * x / (1 + XY)
Find z'y = Z * [ln (1 + XY) + y * x / (1 + XY)]
=(1+xy)^y*[ln(1+xy)+xy/(1+xy)]