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tresset_1 [31]
3 years ago
11

Can someone help me?

Mathematics
1 answer:
nirvana33 [79]3 years ago
6 0
So you can do number one is d
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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}}

6 0
1 year ago
Which angle below is another name for ZABC?
Liula [17]
Answer is none of these
5 0
3 years ago
Solve the equation on the<br> interval [0, 27r).<br> 4(sin x)2 - 2 = 0
Ket [755]

4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)

x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill

Check the picture below.

4 0
2 years ago
1/3(6x-15)=1/2(10x-4) solve for x
Tatiana [17]

Simplify 1/3(6x - 15) to 6x - 15/3

6x - 15/3 = 1/2(10x - 4)

Factor out the common term; 3

3(2x - 5)/3 = 1/2(10x - 4)

Cancel out 3

2x - 5 = 1/2(10x - 4)

Simplify 1/2(10x - 4) to 10x - 4/2

2x - 5 = 10x - 4/2

Factor out the common term; 2

2x - 5 = 2(5x - 2)/2

Cancel out 2

2x - 5 = 5x - 2

Subtract 2x from both sides

-5 = 5x - 2 - 2x

Simplify 5x - 2 - 2x to 3x - 2

-5 = 3x - 2

Add 2 to both sides

-5 + 2 = 3x

Simplify -5 + 2 to -3

-3 = 3x

Divide both sides by 3

- 1 = x

Switch sides

<u>x = -1</u>

8 0
3 years ago
The local newspaper has letters to the editor from 70 people. If this number represents 4% of all the newspaper readers, how man
vredina [299]

Answer:

The newspaper has 1750 readers.

Step-by-step explanation:

Divide 70 by 2: 35.

Then, multiply 35 by 50 which is equal to 1750.

6 0
3 years ago
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