- #1

- 14

- 0

y(x)=C1e^x+(C2+C3x+C4x^2)e-x

2. I tried to come with differential equation but this is it

I can 't seem how to begin

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter pedro123
- Start date

- #1

- 14

- 0

y(x)=C1e^x+(C2+C3x+C4x^2)e-x

2. I tried to come with differential equation but this is it

I can 't seem how to begin

- #2

- 14

- 0

- #3

- 14

- 0

is the the differential equation -y''''-y''=0 im not sure

- #4

- 9

- 0

- #5

- 14

- 0

and alternate fom beign r^4-r^3

making the general solution

y''''-y'''=0

is this alright thanks.

- #6

- 9

- 0

(r-1)(r[itex]^{3}[/itex]+3r[itex]^{2}[/itex]+3r+1)=r[itex]^{4}[/itex]+2r[itex]^{3}[/itex]-2r-1

- #7

- 14

- 0

y''''-2y'''-2y-1=0

is this alright and sorry about my mistake in factoring

- #8

- 9

- 0

y[itex]^{(4)}[/itex]+2y[itex]^{(3)}[/itex]-2y'-y

Remember that r=r[itex]^{1}[/itex] and 1=r[itex]^{0}[/itex]

- #9

- 14

- 0

I have a problem which in involves a second order differential equations with imaginary roots and I can seem to know how to finish the problem.

d^2y/dt^2 +15y =cos 4t+2 sin t

this is what I got so far

r^2+15=0 for the homogeneous part

r=+-(√15)

Yh=C1cos√15+C2sin√15

and now the part that follows is to come with a particular solution for

cos 4t+2 sin t

but I don't know how to properlly set up

- #10

- 14

- 0

and thanks for helping me so far

- #11

- 14

- 0

I know that 2 is a constant but can my particular solution be

Yh=Acos4t +Bsin4t

Yh=Acos4t +Bsin4t

- #12

Mark44

Mentor

- 35,057

- 6,793

Share:

- Replies
- 3

- Views
- 4K

- Replies
- 8

- Views
- 6K

- Replies
- 1

- Views
- 1K