Problems of differential equations The general solution of the differential equation dy / DX + ycosx-e ^ (- SiNx) = 0 is,

Problems of differential equations The general solution of the differential equation dy / DX + ycosx-e ^ (- SiNx) = 0 is,

Remove the one without y, solve it, and then use the constant variation method. Write the process by yourself. It's very annoying

If x and y are positive real numbers and 2x + 3Y = 1, then the minimum value of 1x + 1y is___ ．

∵ 2x + 3Y = 1, ∵ 1x + 1y = (1x + 1y) (2x + 3Y) = 2 + 3YX + 2XY + 3 ∵ x, y is a positive real number, ∵ 3YX + 2XY ≥ 23yx2xy = 26 ∵ 2 + 3YX + 2XY + 3 ≥ 5 + 26 ∵ the minimum value of 1y is 5 + 26, so the answer is 5 + 26

The general solution of calculus equation y '+ y = e ^ - x

Characteristic equation R + 1 = 0
r=-1
So the homogeneous general solution y = CE ^ (- x)
You can see that the right side of the equal sign is in the general solution
So let the special solution be y = axe ^ (- x)
y'=ae^(-x)-axe^(-x)
Substituting into the original equation
ae^(-x)-axe^(-x)+axe^(-x)=e^(-x)
a=1
So the special solution is y = Xe ^ (- x)
The general solution of the equation is
The special solution is y = CE ^ (- x) + Xe ^ (- x)

Calculus: find the general solution of the following first order linear equation. Y '- 2Y / x = x ^ 2

P(x)=-2/x,Q(x)=x^2
∫2/xdx ∫-2/xdx
y=e^ (∫x^2 e dx+c)
y=x^2(∫x^2(1/x^2)dx+c)
y=x^2(x+c)

The calculus equation of Y '' - 2Y '+ y = 0

Solving characteristic equation first
λ²-2λ+1=0
We obtain that λ = 1 is a double root
So the general solution is y = C1E ^ x + C2 Xe ^ X

General solution of equation 2Y '- y = e ^ x

2y'- y = e^x
That is y '- Y / 2 = (e ^ x) / 2
By using the general solution formula of the first-order linear non-homogeneous differential equation, it is obtained that
y = e^(∫1/2 dx)[∫（e^x）/2 * e^(-∫1/2 dx)dx + C]
= e^(x/2)[e^(x/2) + C]
= C e^(x/2) + e^x

The general solution process of the equation y "- 5Y '+ 6y = (x + 1) e ×

1. Find the general solution of the homogeneous equation y '' - 5Y '+ 6y = 0
The characteristic equation is R & # 178; - 5R + 6 = 0
The eigenvalue R1 = 2 and R2 = 3 are obtained
The general solution y = C1 * e ^ 2x + C2 * e ^ 3x is obtained
2. Because r = 1 is not the solution of the characteristic equation, let y * = C3 * e ^ X be substituted into the equation
Get C3 * e ^ x-5c3 * e ^ x + 6c3 * e ^ x = e ^ X
2C3=1 C3=1/2
So the general solution of the equation is y = C1 * e ^ 2x + C2 * e ^ 3x + 1 / 2 * e ^ X

The general solution of Y '' - y '- 6y = 0
Urgent··
All brothers and sisters, help

This problem is called homogeneous differential equation with constant coefficients, which is solved by eigenvalue method
The characteristic equation is x ^ 2-x-6 = 0, and the eigenvalue is x = 3, x = - 2;
So the general solution of the equation is y = C1E ^ (3x) + c2e ^ 2 (- 2x), where C1 and C2 are two arbitrary constants

The general solution of calculus y '+ 2Y' + 5Y = 0

The characteristic equation a ^ 2 + 2A + 5 = 0 has conjugate complex roots - 1 + 2I, - 1-2i
So the general solution is y = e ^ (- x) (c1cos2x + c2sin2x)

General solution of calculus y '= Y2

dy/dx=y²
dy/y²=dx
integral
-1/y=x+C
y=-1/(x+C)