Solution of differential equation: dy / DX – 2Y / (x + 1) = (x + 1) ^ 5 / 2

Solution of differential equation: dy / DX – 2Y / (x + 1) = (x + 1) ^ 5 / 2

Dy / DX – 2Y / (x + 1) = (x + 1) ^ 5 / 2 is a first-order linear equation. The general solution formula is:
y=(x+1)^2(C+∫(x+1)^(1/2)dx)
=(x+1)^2(C+(2/3)(x+1)^(3/2))
=C(x+1)^2+(2/3)(x+1)^(7/2)

Y = cos (XY) - x, find its differential

dy=-sin(xy)dxy-dx
=-sin(xy)xdy-sin(xy)ydx-dx
So dy = - [ysin (XY) + 1] / [xsin (XY) + 1]

Find the differential of the implicit function determined by equation x ^ 2 + XY + y ^ 2 = 3

2xdx+ydx+xdy+2ydy=0
(x+2y)dy=-(2x+y)dx
dy=-(2x+y)/(x+2y) × dx

Let the function y = f (x) be determined by the equation E2x + y-cos (XY) = E-1, then the normal equation of curve y = f (x) at point (0,1) is __

From the problem set, take the derivative of X on both sides of E2x + y-cos (XY) = E-1, and get
e2x+y•[2+y′]+sin（xy）•[y+xy']=0
Substitute x = 0 into the original equation to get y = 1,
Then substitute x = 0 and y = 1 into the above formula to obtain
Y '|x = 0 = - 2. Therefore, the normal equation is
y−1＝1
2(x−0)
Namely   x-2y+2=0．

Let the function y = f (x) be determined by the equation E ^ (2x + y) + cos (XY) = E-1, then dy=_____ Let the function y = f (x) be determined by the equation E ^ (2x + y) - cos (XY) = E-1, then dy = ___. The above is wrong.

=-[ysin(xy)+2e^(2x+y)]/[ysin(xy)+e^(2x+y)]*(dx)

From cos (XY) = x + y, determine that y is a function of X and find y '. Can you write down the process of the problem just now? I don't understand

cos(xy)=x+y
Derivation of X on both sides:
-sin(xy)*(y+xy′)=1+y′
y′=-[1+ysin(xy)]/[xsin(xy)+1]
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Derivation of COS on the left: - sin (XY)
Then take the derivative of XY: y + XY ′
Derivative of X on the right: 1 + y ′
Multiply the two terms on the left and finally solve y '