y y 3 y beg for the next y to the process thank you

Solution: y''=y'^3+y' ==>dy'/(y'^3+y' )=dx

[1/y'-y'/(y'^2+1)]dy'=dx=>ln│y'│-(1/2)ln(y'2+1)=x+ln c1 (c1 is the integration constant).

y'/√(y'^2+1)=c1e^x

y'^2/(y'^2+1)=c1^2e^(2x)=>y'^2=c1^2e^(2x)/[1-c1^2e^(2x)]=>y'= c1e x [1-c1 2e (2x)]y= c1 e xdx [1-c1 2e (2x)] dt (let sint=c1e x, resimplify) c2 t (c2 is the integration constant).

c2±arcsin(c1e^x)

Therefore, the general solution of the original equation is y=c2 arcsin(c1e x) (c1,c2 is the integral constant).

y^3-y^2+y=0

y(y^2-y+1)=0

Because the one in y 2-y+1=0 can't be typed, you should know that that one is less than 0, so y=0

Because of the odd bend silver power, the assignment of y and the output of y-1 can be positive or negative, and it can also be annihilated.

For reference, please smile.

It can be with a negative sign, which refers to the chaos of the value in the hall to open 3 power, Nian Yin inside is (y-1), open the cubic is (y-1), add a negative sign in front, the answer is -(y-1), that is, 1-y.

3. The problem of all inverse functions should be relatively simple, find the cube on both sides of the equation, and then transform the equation so that x is on the left side of the equation, and the second side is 1, it should be able to solve your problem, I hope it can help you, if you don't understand, you can continue to ask, and you can adopt it if you are satisfied, come on.

Replace x with y, and y with x

The cubic of both sides gets:

x³=y+2

Move the term to get: y=x -2

Since the domain of the original function is r, the domain of the inverse function is r.

Lingy'=p, file.

Then y"=dp/dx=dp/dy

dy/dx=p*dp/dy

So the original equation can be reduced to.

p*dp/dy

p=py is dp

y-1)*dy

The integration of both sides of the equation yields p=y'=

y+cc is constant).

x=y=2, y'=1/2

That is, 1 2 = 2 + c is solved to obtain c =

So. y'=

Old. 2dy/(y-1)^2=dx

The two sides of the hidden chakra are divided.

2/(y-1)=x

c and x = y = 2

Substituting to get c=

Therefore, the general solution of the equation is the line carrying liquid.

2/(y-1)=x

i.e. (x-4) (y-1) + 2 = 0

y''-4y'+4y=e 2x.

Corresponds to homogeneous equations.

y''-4y'The eigenequation of +4y=0.

is: r 2-4r + 4 = 0

The characteristic root is: r1=r2=2

General solution: y=(c1+c2x)e 2x

Since r=2 is the double root of the characteristic equation of Cha Sakura, it is not eliminated to let y=ax 2e 2x, then y =2axe 2x+2ax 2e 2xy take the slag=2ae 2x+8axe 2x+4ax 2e 2x substituting the original equation:

2ae^2x+8axe^2x+4ax^2e^2x-4(2axe^2x+2ax^2e^2x)+4ax^2e^2x=e^2x

2ae^2x=e^2x

2a=1 gives a=1 2

Therefore, a special solution is found as: y=

1/2)x^2e^2x

Therefore, the general solution is as follows: y=(c1+c2x)e 2x+1 2)x 2e 2x

where c1, c2 are arbitrary constants).

x²＋y²－4x＋6y＋13＝0,﹙x²－4x＋4﹚＋﹙y²＋6y＋9﹚＝0

x－2﹚²＋y＋3﹚²＝0

Solution: Sou jujube x 2, y 3

Simplify, got. Original (3x y)[(3x y 3 x y] x 9y

3x＋y﹚﹙﹣2y﹚－x²＋9y²

Shiji split 6xy 2y x 9y

6xy＋7y²－x²

Substitute x 2, y 3 into the above formula, get.

Sharp 36 63 4

Solution x 2 + y 2 + 2 x + 4y - 3 = 0

x+1）²+y+2）²=8

The distance from the center of the circle (-1, -2) to the straight is (-1-2+1) 2= 2 radius first.

Therefore, the circle intersects the straight line (it is better to draw a sketch for easy observation), so there are 4 points with a distance of 2 2.