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ivolga24 [154]
3 years ago
14

What is the value of x in the equation

Mathematics
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

54

Step-by-step explanation:

\frac{1}{3x}  -  \frac{1}{2}  = 18 \frac{1}{2}  \\  \\  \frac{1}{3x}  -  \frac{1}{2}  =  \frac{37}{2}   \\  \\  \frac{1}{3x}  =  \frac{37}{2}  -  \frac{1}{2}  \\  \\  \frac{1}{3x}  =  \frac{37 - 1}{2}  \\  \\  \frac{1}{3x}  =  \frac{36}{2}  \\  \\  \frac{1}{3x}  = 18 \\  \\ 1 = 54x

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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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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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