The general solution of Y '' + 2Y '+ 10Y = 0 differential equation

The general solution of Y '' + 2Y '+ 10Y = 0 differential equation

The original equation is a second order homogeneous linear differential equation with constant coefficients,
Solve the characteristic equation R ^ 2 + 2R + 10 = 0,
The characteristic roots R1 = - 1 + 3I, R2 = - 1-3i are obtained,
So the general solution of the original differential equation is
y=e^(-x)(C1cos3x+C2sin3x).

The general solution of the differential equation is y = CX ^ 2 + X____

y = Cx^2 + x (1)
y' = 2Cx+1 (2)
y'' = 2C (3)
from (2)
(y')^2 = 4C^2x^2+ 4Cx + 1
= 4C(Cx^2+x) +1
= 2y''y+1
The differential equation with general solution CX ^ 2 + X is
2y''y-(y')^2 +1 =0

The general solution of the differential equation y '= (x + y) ^ 2 is_____ Solving by implicit function

Let u = x + y
Then u '= 1 + y' = = > y '= u' - 1
There is y '= u ^ 2 = u' - 1
u'=u^2+1
du/(u^2+1)=dx
arctanu=x+c
u=tan(x+c)
x+y=tan(x+c)
So y = Tan (x + C) - X

First order linear differential equation y '= x / y + Y / X

Let y = XT, then y '= t + XDT / DX
By substituting into the original equation, XDT / DX = 1 / T
==>tdt=dx/x
==>T & sup2 / / 2 + ln │ C │ = ln │ x │ (C is an integral constant)
==>y²/(2x²)+ln│C│=ln│x│
==>Ce^(y²/(2x²))=x
So the general solution of the original differential equation is x = CE ^ (Y & sup2; / (2x & sup2;)) (C is an integral constant)

It is known that y = 1, y = x, y = x & # 178; are three solutions of a second order non-homogeneous linear differential equation?

The general solution of the equation
y=C1(x²-1)+C2(x-1)+1

Find the general solution of the differential equation y '+ 2XY = 2xe (- x2)

Y '+ 2XY = 2xe ^ (- x ^ 2) y' + 2XY = 0y '/ y = - 2x (LNY)' = - 2xlny = - x ^ 2 + c0y = CE ^ (- x ^ 2) let y = C (x) e ^ (- x ^ 2) y '= C' (x) e ^ (- x ^ 2) + (- 2x) * C (x) e ^ (- x ^ 2) = 2xe ^ (- x ^ 2) C '(x) = 2XC (x) = ∫ 2xdx = x ^ 2 + C1 general solution y = (x ^ 2 + C1) e ^ (- x ^ 2)

For the general solution of the differential equation y '- 2XY = 2xe ^ (x ^ 2), please write the calculation process

This is a non-homogeneous first-order linear differential equation. First, we find the corresponding homogeneous linear equation y '- 2XY = 0, dy / DX = 2XY, dy / y = 2xdx, ∫ dy / y = ∫ 2xdx, LNY + C1 = x & # 178; + C2, y = CE ^ (X & # 178;). By using the constant variation method, we change C into u, that is, let y = UE ^ (X & # 178;) then dy / DX = u'e ^ (X & # 178;)

The second derivative of y = the square of 1 + (the first derivative of Y)

y"=1+y'
Let y '= t, then y "= t',
The original formula is t '= 1 + t,
I'm lazy. I forgot to find the formula of first order differential equation,
T (x) is solved,
And then y '= t (x),
It is also a first order differential equation with formula
Conclusion:
This differential equation does not contain X,
Through the above substitution, the second order differential equation is transformed into two first order differential equations

Find the general solution of the second derivative of the differential equation y + the first derivative of y = the square of X
y’= ax3 + bx2 + cx + d
Why is it in this form?

D & sup2; Y / DX & sup2; + dy / DX = x & sup2; solution: D & sup2; Y / DX & sup2; + dy / DX = 0 the characteristic equation is λ & sup2; + λ = 0, λ (λ + 1) = 0, λ & # 8321; = 0, λ & # 8322; = - 1

The square of the second derivative of Y minus the first derivative of a times y is equal to zero, and the general solution of the differential equation is obtained

y"-ay'^2=0
y"/y'=ay'
Points:
lny'+C1=ay+C2
lny'=ay+C
y'=Ce^(ay)
y'e^(-ay)=C
Points:
e^(-ay)=-Cx/a
-Ay = lnx-c '(C' is a constant)