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Sever21 [200]
3 years ago
6

Help plz 15 points!!!!!!!!!!!!!!!! really inportant

Mathematics
2 answers:
fiasKO [112]3 years ago
5 0
Question 14) 3³


question 15) 100/9
Angelina_Jolie [31]3 years ago
3 0
14. C
15. C

Hope it helpsssss!!!
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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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Numbers in a problem or related numbers that are easy to work with mentally
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Find the product of<br> 2x^2(x + 3)
KengaRu [80]

Answer:

16x

Step-by-step explanation:

2x^2(x+3)

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4x(x+3)

4x + 12x =

16x

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Given: AB=CD, BD=DE<br> Prove: AD=CE
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This question is incomplete. Can you maybe add more information?

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