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The general solution of a non-homogeneous linear differential equation is equal to the general solution of its corresponding homogeneous differential equation plus a special solution of the non-homogeneous differential equation.

The homogeneous equation of the second-order constant coefficients.

y''+py'+qy=0

Characteristic equations. r²+pr+q=0

r1,r2=[-p (p -4q)] 2 is a real root. R1≠R2 Differential equation solution y=C1E (R1X)+C2E (R2X).

r1=r2 differential equation solution y=x[c1e (r1x)+c2]r1,r2= i differential equation solution y=e ( x)(c1cos x+c2sin x).

The second-order constant coefficient is a non-homogeneous equation.

y''+py'+qy=f(x)

f(x)=e ( x)p(x) p(x) is the polynomial with the highest order m.

The form of the solution is y=y+y*

y*=(x^k)e^(λx)q(x) q(x)=a0x^m+a1x^(m-1)+.am

Not the root of the eigenequation (≠r1 ≠r2) k=0 is the single root of the eigenequation (=r1 or =r2) k=1 is the double root of the eigenequation (=r1=r2) k=2

y*''+py*'+qy*=e ( x)p(x) each order coefficient is equal to the solution a1,a2,..am derives y*f''(u)=4f(u)+u

r²=4 r1=-2 r2=2

y=c1e^(-2x)+c2e^(2x)

y*=a0u+a1

y*'=a0

y*''=0

y*''= 4y*+u

0=4(a0u+a1)+u

a0=-1/4

a1=0y=c1e^(-2x)+c2e^(2x)-u/4

Further Mathematics. College courses) Calculus.

University Courses Courses Mathematics Differential Equations.

Ability to separate variables. xy'+1=4e (-y)(dy dx)x=(4-e y) e y [e y (e y-4)]dy=(-1 crack dust x)dx ln(e y -4)=-lnx +c=ln(e c x) e y=(4x+e c) x x detongfinch excavation: y=ln[(4x+e c) x].

Known curves, find.

1) the tangent equation on the curve when x=1;

2) Find the area of the plane figure enclosed by the curved ridge line, the tangent line and the x-axis, and the volume of the rotating body formed by its rotation around the x-axis.

Variable upper and lower limits derivative formula.

f(b(x))b'(x)-f(a(x))a'(x) So bring in to get.

2xsin(x 4)-1 2 sin(x) root number x

There is obviously something wrong with this topic, and 0 can't get an equal sign at all.

The first two have equal limit values at 0, and the last two have limit values at 1, one is not sin1, and the other is 1, so the function is continuous at 0 and discontinuous at 1.

f(x)

ln(1+x)/x ; x<0=sinx/x ; 0≤ x≤1=x^2 ;x>1f(0-) = lim(x->0) ln(1+x) x =1f(0) is not defined.

f(0+) lim(x->0) sin x =1x=0 , f(x) discontinuous.

f(1)=f(1-) =lim(x->1) sinx/x = sin1/1

f(1+) = lim(x->1) x 2 =1x=1, f(x) is discontinuous.

ans :d

According to the integral median theorem, in (0,1) there exists a number a such that f(a) is equal to that integral, and f'(x) >0, saying that the function increases on [0,1], and a<1, so f(a).

According to the derivative greater than zero, it can be seen that the function is monotonically increasing in the range from 0 to 1, so the function value is maximized when x=1. The integral of the function in the range of 0 to 1 is equal to the area of the function image enclosed by the x-axis y-axis, and the average height of this area is less than the value of the function when x is taken at one time, and the width of this area is equal to one, then this area must be less than the value of the function when x is taken at one time.

Using the method of function transformation, the inverse transformation of u=x-y,v=x+y is x=(v+u) 2 y=(v-u) 2 The value of the function determinant of the transformation is j=1 2 The transformed region is d1= (in fact, the original region is rotated 45 degrees counterclockwise and then the distance of each point from the center is extended to 2 times the original root number) Therefore, the original integral formula is equal to i= (d1)e v*(1 2)dudv =(1 2) [1,1]du [-1,1]e vdv =e-1 e

The planar sheet with areal density of (x, y)=x is determined by x -1 y 1-x , (x 0) m, and the quality of the sheet is tried.

Solution: Let the mass be m, then:

According to the BAI integral formula:

x^dun*e^(ax)dx = 1/a*x^zhin*e^(ax) -n/a∫x^(n-1)*e^(ax)dx

take, n=3, a=-, substituting the above formula, get dao to:

x^3*e^(-x)dx = -1/λ*x^3*e^(-x) +3/λ∫x^2*e^(-x)dx

1/λ*x^3*e^(-x) +3/λ*[1/λ*x^2*e^(-x) +2/λ∫x*e^(-x)dx]

1/λ*x^3*e^(-x) +3/λ*

-1/λ*x^3-3/λ^2*x^2-6/λ^3*x-6/λ^4)*e^(-x)

0 (x-->6/λ^4） (x-->0)

That is, the generalized definite integral of the equation is 6 4.

The above is the answer, the gamma function formula, I hope it can help you.