Derivative problems. y 2y y 5 f t y when t 2 y 3 find when t 1 y ?

dy/dt=-2y(y+5)

Separate variables. Dy [y(y+5)]=-2dt on both sides. dy[y(y+5)]=-2 dt. 1/y(y+5)

1/5)[1/y-1/(y+5)]

dy/[y(y+5)]

1/5)[∫dy/y-∫dy/(y+5)](1/5)[lny-ln(y+5)]

2∫dt=-2t+c

lny-ln(y+5)=10t+c

t=2, y=3 substitution.

ln3-ln8=20+c

c=ln(3/8)-20

t=1ln(y/(y+5))=10*1+ln(3/8)-20y/(y+5)=exp(ln(3/8)-10)=(3/8)exp(-10)

y=(3/8)exp(-10)(y+5)

y=(15/8)exp(-10)/[1-(3/8)exp(-10)]15/[8exp(10)-3]

Summary. Let the function f(x) be determined by the equation e"=1-2y OK, find dy

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y''=1-2y

y''+2y=1

The characteristic equation or r 2+2=0, r1 = 2i, r2=- 2i is the general solution of the homogeneous equation y=c1*cos( 2x) +c2*sin( 2x), where c1 and c2 are arbitrary constant modulo collapsed branches.

Since the non-semi-seminal order term of the original equation is the 0th order polynomial of x, its special solution form is y*=a, then y*''0

0+2a=1, a=1 2, that is, a special solution of the original equation y*=1 2, so the general solution of the original equation y=y+y*=c1*cos( 2x) + c2*sin( 2x) +1 2

z=(1+xy)^y

lnz=yln(1+xy)

Derivative of y. 1/z)*z'Nucleus = ln(1+xy)+y*1 (1+xy)*(1+xy).'

ln(1+xy)+xy/(1+xy)

So z'Punkaidu=(1+xy) y*[ln(1+xy)+xy(1+xy)]

x=y=1 is substituted for the grandson.

So the original formula = 2ln2+1

Summary. Okay ha

Let y=e x (1+x) and find the derivative of y.

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Hello! This is a composite function derivation problem

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y= 2+3 Find the derivative of the following function.

1) y = (2x +3) (3x-1) solution: y = (2x +3) (3x-1) = 6x -2x +9x-3, so y = 18x -4x + 9 (2) y = (x-2) solution: y = (x-2) =x-2 when x 2, y = x-2, at this time y = 1;When x 2, y=-(x-2)=-x+2, and y =-1

3) y=x-sin(x 2)cos(x 2)solution:y Shila=x-=1-=1-[(1 2)cos (x 2)-(1 2)sin (x 2)]=1-(1 2)[cos (x 2)-sin (x 2)]=1-(1 2)cosxNote"Why (3x)'=3?"It is because the derivative of the constant is (3x) =3 x+3x =0 x+3 1=3, i.e., c = 0, 3 is constant, so 3 = 0;(x) nx search and scatter slip

The method is as follows, please comma circle for reference:

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