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The general solution of the second derivative of Y + the first derivative of 2Y - 3Y = 6x + 1 The corresponding homogeneous equation is y '' + 2Y '- 3Y = 0, and the characteristic equation is γ^ 2+2 γ- 3 = 0, the general solution is y = C1E ^ x + c2e ^ (- 3x). Since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1. Therefore, a = - 2, B = - 5 / 3, y * = - 2x-5 / 3, then the general solution of the original equation is y = C1E ^ x + c2e ^ (- 3x) + 6 --- 2x 5 / 3 What I don't understand is "since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1, so what does it mean by a = - 2, B = - 5 / 3, y * = - 2x-5 / 3" What 0 is not the root of the characteristic equation? What is 0? How did ax + B come from? How did you bring the general solution y * = ax + B into the equation? What should a and B do? Please give me your advice. I can't understand it
Write e ^ (0x) (6x + 1) on the right
Judge that the right index 0 is not a root, so the special solution y is the same as 6x + 1, that is, y = ax + B, and then substitute y '' + 2Y '- 3Y = 6x + 1 to find a and B
If the problem is y '' + 2Y '= 6x + 1, then the index 0 on the right is the root, so the special solution y is the same as X (6x + 1), that is, y = x (AX + b)
If the problem is y '' = 6x + 1, then the index 0 on the right is a double root, so the special solution y is the same as x ^ 2 (6x + 1), that is, y = x ^ 2 (AX + b)
The general solution of the second derivative of Y + the first derivative of 2Y - 3Y = 0
Answer: y = C З e ^ (- 3x) + C З e ^ x____________________________________________________________________ Characteristic equation: λ² + two λ - 3 = 0 λ = [- 2 ± √(4 - 4(- 3))]/2= (- 2 ± 4)/2= -3 or 1y₁...
The third derivative of y = the second derivative of Y, find the general solution y Such as title
Third derivative of y = second derivative of Y
Let the second derivative of y be Z
That is, the derivative of Z = Z
So z = e ^ x + C
That is, the second derivative of y = e ^ x + C
So y = e ^ x + ax ^ 2 + BX + C
a. B, C are arbitrary constants
Let x ^ 2Y + XY ^ 2 + 2Y ^ 3 = 1 determine y = y (x), then the derivative of Y is?
Both sides of the equation can be derived at the same time: (this is a general method for deriving implicit functions)
2xy+y'*x^2+y^2+2xy*y'+6y*y'=0
--> y'= -(2xy+y^2)/(x^2+2xy+6y)
I hope I can help you
Max u = (X-10) ^ 0.6 * (Y-5) ^ 0.4 st.3x + 2Y = 100 how to find the derivative
3x+2y=100 ->y=(100-3x)/2
max u=（x-10)^0.6*((100-3x)/2-5)^0.4
=(x-10)^0.6*(45-1.5x)^0.4
u'=0.6(x-10)^(-0.4)*(45-1.5x)^0.4+(x-10)^0.6*0.4*(45-1.5x)^(-0.6)*(-1.5)=0
=>(45-1.5x)/(x-10)=1
=>x=22
y=17
(x / x + y + 2Y / x + y) multiplied by XY / x + 2Y divided by (1 / x + 1 / y) (2) (3x ^ 2 / 4Y) ^ 2 multiplied by 2Y / 3x + x ^ 2 / 2Y ^ 2 divided by 2Y ^ 2 / X
(x / x + y + 2Y / x + y) multiplied by XY / x + 2Y divided by (1 / x + 1 / y) = (x + 2Y) / [(x + y) / [(x + y) / (XY)] = XY / (x + y) = (XY) ^ 2 / (x + y) ^ 2 (3x ^ 2 / 4Y) ^ 2 multiplied by 2Y / 3x + x ^ 2 / 2Y ^ 2 divided by 2Y ^ 2 / x = 9x ^ 4 / 16y ^ 2 * 2Y / 3x + x ^ 2 / 2Y ^ 2 = 3x ^ 3 / 8y + x ^ 3 / 4Y ^ 4
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