The general solution of the high number y y y 3y 3x 1... There are 3 questions in total

1. The general solution of the differential equation is y"-2y'-3y=0. The characteristic equation is: r 2-2r-3 = 0.

Find the two eigenroots, x1=3 and x2=-1, then the general solution is:

y=c1*e (3x)+c2*e (-1x), where c1 and c2 are constants.

2. Separate the variables first: (2x+1)dx=[2y (y 2+1)]dy=[d(y 2+1)] (y 2+1).

Integrate both sides at the same time, yielding: x 2+x=ln(y 2+1)+c; (c is a constant).

Substituting y(0)=0 yields: c=0, i.e., x 2+x=ln(y 2+1).

Therefore, the special solution is: ln(y 2+1)=x 2+x

3. Establish a Cartesian coordinate system O-XYZ with the center of the ball as the origin point O, and the upper and lower surfaces of the attached cuboid are parallel to the plane XOY, the front and rear surfaces are parallel to the plane XOZ, and the left and right surfaces are parallel to the plane YOZ. Let the coordinates of any vertex of the box be (x,y,z), then the symmetry can be used to find the coordinates of the remaining vertices. Finally, the volume of the box is obtained:

v=2x*2y*2z=8xyz。

and (x,y,z) are the vertices of the box, then: x 2 + y 2 + z 2 = a 2.

From the mean inequality, we know that 3*((x 2)*(y 2)*(z 2)) 1 3)<=a 2 (take the equality condition as x=y=z).

So the volume is: v=8xyz<=8(a3) (3 3).

When x=y=z=a3, an equal sign is taken.

At this point, the cuboid is a cube with an edge length of 2a 3 and reaches its maximum volume of 8 (a 3) (3 3).

1, the question is r 2-r-3 = 0 (whether the question is copied wrong, is it minus 2) then that) = 0 is not a feature root.

y=ax+b brings back -a-3 (ax+b)=3x+1a=-1,b=0

y=c1e^r1x+c2e^r2x-x

2, the original formula can be transformed into (2x+1)dx=dy 2 (y 2+1)x 2+x+c=ln(y 2+1)Substituting x=y=0 into c=03, for convenience to remember that the radius is r, abc is length, width and height, (a 2) 2+(b 2) 2+(c 2) 2=r 2

The goal is to have the largest ABC.

The partial derivative of ABCT is equal to zero by using the Lagrangian function l=abc+t[(A 2) 2+(b 2) 2+(c 2) 2-r 2] to get a=b=c=2r (3).

It's so hard to type!

The three questions should be: 2 3a 3

The other 2 questions have formulas, I forgot, I can't be silent, lx, it's up to you, I'll study it too.

Differential equations can be made using the differential operator method.

The characteristic equation r 2r 3 = 0

r1=3，r2=–1

The general solution of the homogeneous equation is y=c1·e ( x) c2·e (3x) to find the special solution of the original equation.

Method 1 (need to master): Let the special solution be y=ax b, then y'=a，y''=0, substituting the original equation yields 3ax 2a 3b=3x 1

3a=3，–2a–3b=1

It can be solved a= 1, b=1 3

The special solution is y= x 1 3

Method 2: You can use the differential operator method (this method is more simple in some complex problems, and can be used as a method of calculation, only for understanding).

d²y–2dy–3y=3x＋1

Solution: y=1 (d 2d 3) (3x 1) ( 1 3 2 9 d ) (3x 1).

1/3 (3x＋1)＋2/9 · 3

x 1 3 so the general solution of the original equation y=c1·e (3x) c2·e ( x) x 1 3

If you have any questions, please feel free to ask.

y’=3y+x

dy-3ydx=xdx

e (-3x)dy-3ye (-3x)dx=xe (-3x)dx (both ends of the equation are in the same vehicle and the noise is e(-3x)).

d(ye (-3x))=d(-(x 3+1 9)e (-3x))=>ye (-3x)=c-(x 3+1 9)e (-3x) (c is the integration constant).

y=ce^(3x)-x/3-1/9

The general solution of the original equation is y=ce (3x)-x 3-1 9

y'=3y/(1+x)

Then 1 3ydy=1 (1+x)dx

Each attendant takes the lead from the world and gets the old points separately.

1/3lny=ln(1+x)+c1

Rule. y=c(1+x)^3

Lingy'=p, then y''=dp/dx=dp/dy*dy/dx=pdp/dy

Bring in the equations. pdp/dy=(y+1)p

Separate the variables and integrate them separately.

dp=∫(y+1)dyp=1/2y^2+y+c1/2

That is: dy dx=1 2y 2+y+c1 2

Separate the variables and integrate them separately.

dy/(1/2y^2+y+c1/2)=∫dx

dy/((y+1)^2+c1-1)=∫2dx

When c1>1, the integral is.

1/√(c1-1)∫dy/([(y+1)/√(c1-1)]^2+1)=∫2dx

1/√(c1-1)*arctan[(y+1)/√(c1-1)]=2x+c2

When c1>1, the integral is.

dy/((y+1+√(1-c1))(y+1)-√1-c1)))=∫2dx

1/(2√(1-c1)*ln|(y+1-√(1-c1))/(y+1+√(1-c1))|=2x+c2

When c1=1, dy(y+1)2=2dx

Then -1 (y+1)=2x+c2

Used: Separation of variables.

Both sides accumulate rubber points at the same time. i.e.: in the round of y'=3x both sides at the same time to get y=(3 Laru brother 2) x 2 hopes.