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ololo11 [35]
2 years ago
15

What is slope of the line

Mathematics
2 answers:
uranmaximum [27]2 years ago
5 0

Answer:

The Answer is 3

Step-by-step explanation:

First you'd want to find two points so example (0,1), and (2,7)

Second You would put them in the slope formula y2-y1/x2-x1

Third You would want to solve the formula by getting 2-0 and 7-1 And get 6/2

Fourth Simply 6/2 and get 3/1 which is also 3.

Hope this helps

aleksley [76]2 years ago
3 0
3 is the correct slope of the line:)
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=  >  \:  - 10x - 7y =  - 12 \\  =  >  \:  - 10(4) - 7y =  - 12 \\  =  >  \:  - 40 - 7y =  - 12 \\  =  >  \:  - 7y =  - 12 + 40 \\  =  >  \:  - 7y = 28 \\  =  > y =   -  \frac{28}{7}  \\  =  >  \: y =  - 4
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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

The characteristic equation

r^2 - 3r + 2 = (r - 1) (r - 2) = 0

has roots at r=1 and r=2. Then the characteristic solution is

y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

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