Question

Teach me the simplest differential equations (second order)

Answer 1

You know what an auxillary equation and general solution is right?

Answer 2

Yes, but I don't know how to convert from auxiliary equations to general solution

Answer 3

I have a few nice books if you want

Answer 4

@zzr0ck3r no thanks

Answer 5

Okay, the general solution when you have two roots (r1 and r2)
it is :
\[y = c_1e^{r_1x} + c_2 e^{r_2x}\]

Answer 6

Let's say you have
y'' + 5y' + 6y = 0
Convert it to an auxillary equation and find the roots.

Answer 7

aight gn all, have fun

Answer 8

r^2+5r+6=0, (r+2)(r+3)=0, r=-2 or r=-3, \(\Large y=ae^{-2x}+be^{-3x}\)
Am I correct?

Answer 9

Yes.

Answer 10

Thanks, that's all?

Answer 11

No, there are two other types of the general solutions

Answer 12

When you have only one root like
y'' + 4y' + 4y = 0
r = -2

The formula is:
\[y = c_1e^{r_1x} + c_2xe^{r_1x}\]

Answer 13

I see

Answer 14

Now the other formula is for when you have complex roots..
\[y = e^{\alpha x}(\cos \beta x + \sin \beta x)\]
where alpha = -b/2a
and \[\beta = \frac{ \sqrt{4ac-b^2} }{ 2a }\]

Answer 15

I mean*
\[y = e^{\alpha x} (c_1\cos \beta x + c_2 \sin \beta x)\]
forgot the constants ;p

Answer 16

I see

Answer 17

Now, this is just evaluating the general solution..
In the chapter where you do face second-order diff equations you also have to solve initial value problems and boundary value problems.

Answer 18

you mean when the root is \(a\pm bi\), alpha is a and beta is b, am i correct

Answer 19

yeah when the root is that

Answer 20

Can you give me a problem? :)

Answer 21

yes alpha is a , and beta is b.

Answer 22

sure:
\[y'' - 6y' + 8y = 0\]

Answer 23

r^2-6r+8=0, (r-2)(r-4)=0, r=2 or r=4
\[\large y=ae^{2x}+be^{4x}\]
am i correct

Answer 24

yep

Answer 25

Alright try this one:
4y'' + y' = 0

Answer 26

4r^2+1=0, r^2=-0.25, r=±0.5i
\[\large y=a\cos0.5x+b\sin0.5x\]
Am I correct

Answer 27

That's a y' not y
remember
y'' = r^2 , y' = r , y = 1

Answer 28

oh sorry i misred it

Answer 29

4r^2+r=0, r(r+4)=0, r=0 or r=-4
\[\large y=a+be^{-4x}\] am i correct?

Answer 30

but if it were 4y'' + y = 0
you would be correct

Answer 31

and yes that is correct

Answer 32

good job!

Answer 33

want to try an initial-value problem?

Answer 34

thanks

Answer 35

yes plz

Answer 36

Try:
4y'' - 4y' + y = 0
where y(0) = 1
and y'(0) = -1.5

Answer 37

Basically in an initial value problem you find the constants after you get the general solution.
you find the constants from evaluating..
in this case you evaluate the general solution at y(0) = 1
and the derivative of it at y'(0) = -1.5

Answer 38

4r^2-4r+1=0, (2r-1)^2=0, r=0.5
\[\large y=ae^{0.5x}+bxe^{0.5x}\]
y(0)=1,
a=1

y'=0.5ae^(0.5x)+be^(0.5x)+0.5bxe^(0.5x)
y'(0)=1,
0.5a+b=1
b=0.5

\[\large\therefore y=e^{0.5x}+0.5xe^{0.5x}\]
Am I correct

Answer 39

you got the first constant right the second one is wrong

Answer 40

\[y'(x) = (1/2)e^{1/2x} + c_2 (e^{x/2} +1/2xe^{1/2})\]
y'(0) = -1.5
not 1

Answer 41

oh careless me

Answer 42

y'(0)=-1.5
0.5a+b=-1.5
b=-2

\[\large\therefore y=e^{0.5x}-2xe^{0.5x}\]
Am i correct?

Answer 43

yes, you are right now

Answer 44

Wow thanks a lot :D Wish I could give you a million medals... :)

Answer 45

lol, no problem :P