Can someone help me?

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

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