The process of solving the general solution of the differential equation y 2y y x requires speed

Solution: The corresponding homogeneous equation is y''-2y'+y=0, the eigenequation is r 2-2r+1=0, there is a real root r=1, so the general solution of the homogeneous equation corresponding to the given equation is .

y=(c1+c2x)e^x

Since =0 is not the root of the eigenequation, we should let y*=b0x+b1, then b0x-2b0+b1=x,> b0=1,b1=2,> y*=x+2, and the general solution is y=(c1+c2x)e x+x+2

Homogeneous characteristic equation: r 2-2+1=0

r1=r2=1

Its homogeneous general solution: y=(c1+c2x)e x

It is observed that y=x+2

Therefore its solution is y=(c1+c2x)e x+x +2

First, find the solution of the homogeneous equation:

The characteristic equation is r -2r+1=0

r1=r2=1

The general solution of the homogeneous equation is y=e x(c1+c2x) and then find the special solution of the non-secondary equation:

The special solution form is y0=e ( x)·x k· (ax+b)p(x)=x=e ( x)·x

0 is not the root of the eigenequation.

k=0y0=e^(0x)·x^0·(ax+b)=ax+by′=a,y″=0

Bring in the original equation: 0-2a+ax+b=x

a=1,b=2

y0=x+2

The solution of the original equation is y=e x(c1+c2x)+x+2.

I hope my answer is helpful to you.

Multiply both sides by e (x 2) and note (e (x 2) y).'=(y'+2xy)e^(x^2)

I get the points on both sides.

e^(x^2)y)=x^2+c

So y=x 2e (-x 2)+ce (-x 2).

The characteristic equation a 2 +2a + 5 = 0 has a conjugated spine and a releasing beam seepage root -1+2i, -1-2i

Therefore, the general solution of slag destruction is y=e (-x) (c1cos2x+c2sin2x).

Set y'=p, then y''=dp dy*dy dx=pdp dypdp dy=(1+p 2) 2y

2pdp/(1+p^2)=dy/y

ln(1+p^2)=ln|y|+ln|c|Get 1+p 2=cy

y'= Punch ruler (cy-1).

dy/√(cy-1)=dx

2 c (cy-1) = x+c'(c,c'The scatter height is constant).

1) First, find the general solution of the homogeneous differential equation y''+3y'+2y=0:

The characteristic equation is r 2 + 3r + 2 = 0, and the solution is r1=-1, r2=-2, so the general solution of the homogeneous differential equation is y h=c 1e +c 2e (c 1 and c 2 are any constants of the honor line).

Next, find the special solution of the non-homogeneous differential equation y''+3y'+2y=3x+1:

Assuming that the special solution is y p=ax+b, substituting the original equation gives 2a+3a+2ax+2b=3x+1

Comparing the coefficients of the same power, we get 2a+3b=1, 2a=3, and the solution is a=3 2, b=-5 4

Therefore, the general solution of this non-homogeneous differential equation is y=c 1e +c 2e +3 2x-5 4

2) First, find the general solution of the homogeneous differential equation y''+y'=0:

The eigenequation is r 2 + r = 0, and the solution is r 1 = 0 and r 2 = -1, so the general solution of the homogeneous differential equation is y h=c 1 + c 2e (c 1 and c 2 are arbitrary constants).

Next, find the special solution of the non-homogeneous differential equation y''-5y'+y'=x:

First, find the general solution of the homogeneous differential equation y''-5y'+y'=0:

The characteristic equation is r 2-5r+1=0, and the solution is r1=(5+ 21) 2, r2=(5- 21) 2, so the general solution of the homogeneous differential equation is y h=c 1e +c 2e (c 1, c 2 are arbitrary constants).

Since the non-homogeneous term is a one-time function, we guess that its special solution is y p=ax+b, and substituting the original equation gives a+bx-5ax+2a=12x (21-21).

Comparing the coefficients of the same power, we get -a+2a=12 (21-21), 1-5a+b=0, and a=12 (21-21) 3, b=5a-1=-(11+ 21) (21-21).

Therefore, the general solution of this nonhomogeneous differential equation is y=c 1+c 2e +12 (21- 21)x-(11+ 21) (21- 21).

y''/y'^2=2y/(y^2+1)(-1/y')'Dan = (ln(y 2+1)).'Integral chain light on both sides: -1 y'=ln(y 2+1)+c1-dx=(ln(y 2+1)+c1)dy: -x= ln(y 2+1)dy+c1y=yln(y 2+1)- y*2y (y 2+1)dy+c1y=yln(y 2+1)- 2y 2+2-1) (y 2+1)dy+c1y=yln(y 2+1)-

Solution: The corresponding homogeneous equation is.

y''-2y'+y=0, and the characteristic equation is.

r 2-2r+1=0, there is a real root r=1, so the general solution of the homogeneous equation corresponding to the given equation is.

y=(c1+c2x)e^x

Since =0 is not the root of the eigenequation, y*=b0x+b1, then.

b0x-2b0+b1=x,=>

b0=1,b1=2,=>

y*=x+2, the general solution is .

y=(c1+c2x)e^x+x+2

y"y'2y

xy"y'2y

According to the theory of differential equations, the general solution of (1) is the sum of a special solution of (1) and the general solution of (2).

1) Special solution: y

ax+b brings in (1) to solve a,b.

2) General solution: y*

c₁e^(s₁x)

c₂e^(s₂x)

s₁=2,s₂=-1.

1) General solution: y

y₁y*ax+b

c₁e^(s₁x)

c₂e^(s₂x)

Homogeneous characteristic equation: r 2-2+1=0

r1=r2=1

Its homogeneous general solution: y=(c1+c2x)e x

It is observed that y=x+2

Therefore its solution is y=(c1+c2x)e x+x

First, find the solution of the homogeneous equation:

The characteristic equation is r -2r+1=0

r1=r2=1

The general solution of the homogeneous equation is y=e x(c1+c2x) and then find the special solution of the non-secondary equation:

The special solution form is y0=e ( x)·x k· (ax+b)p(x)=x=e ( x)·x

0 is not the root of the eigenequation.

k=0∴y0=e^(0x)·x^0·(ax+b)=ax+b∴y′=a,y″=0

Bring in the original equation: 0-2a+ax+b=x

a=1,b=2

y0=x+2

The solution of the original equation is y=e x(c1+c2x)+x+2.

I hope my answer is helpful to you.

dy/dx=2y/x

Just use the separation variable method.

dy/2y=dx/x

The integration of both sides yields: (lny) 2=lnx+c

lny=2lnx+2c

y=e^(2lnx+2c)=(e^2c)*e^(2lnx)=(e^2c)x^2

c is an arbitrary constant, so e 2c is a constant.

y=ax 2 is the general solution.