General solution of differential equation DX / dy equal to 1 / X

xdx==dy
By integrating both sides at the same time, we can get:
1/2*x^2+c1==y+c2
y==1/2*x^2+c

How to find the indefinite integral of dy / y ^ 2-1

1/(y^2-1) =1/(y-1)(y+1)=1/2(1/(y-1)-1/(y+1))
How to find the indefinite fraction of dy / (y + 1), do you know

-Y / (y ^ 2 + 1) dy indefinite integral

Seeking indefinite integral ∫ [(y ^ 2) - 1] / [(y ^ 3) - 3Y] ^ 2]

∫[(y^2)-1]/[(y^3)-3y]^2dy=∫[d(y^3-3y)/3]/[(y^3)-3y]^2dy=-(1/3)/(Y^3-3Y)

RELATED INFORMATIONS

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18. The general solution formula of the first order linear differential equation (X-2) * dy / DX = y + 2 * (X-2) ^ 3, find the general solution of Y ∵(x-2)*dy/dx=y+2*(x-2)³ ==>(x-2)dy=[y+2*(x-2)³]dx ==>(x-2)dy-ydx=2*(x-2)³dx ==>[(x-2)dy-ydx]/(x-2)²=2*(x-2)dx ==>d[y/(x-2)]=d[(x-2)²] ==>y/(x-2)=(x-2)² +C & nbsp; (C is an integral constant) & nbsp; & nbsp; & nbsp; = = & gt; y = (X-2) & # 179; + C (X-2) & nbsp; & nbsp; & nbsp;; the general solution of the original equation is y = (X-2) & # 179; + C (X-2) & nbsp; (C is an integral constant) How to get the following from the above formula?
19. The solution dy / DX + 1 / 3Y = 1 / 3 * (1-2x) y ^ 4 I feel the answer is wrong,
20. Try to derive from DX / dy = 1 / y ': D ^ 2x / dy ^ 2 = - y' '/ (y') ^ 3 the problem of D [1 / y '] / DX} * [DX / dy] D [1 / y '] / DX} * [DX / dy] (because y' in 1 / y 'is the derivative of the function y = f (x), it is the function of X, so 1 / y' is also the function of X. if the function of X wants to derive y now, it needs the derivative method of composite function. For 1 / y ', it needs to derive x first, and then y) Since y is a function of X, why does 1 / y 'need to derive x from y after deriving x? Is it because the inverse function x is also a function of Y A little dizzy