Question

solve the following equation:
y"-4y'+13y=0
y(0)=-1
y'(0)=2

Answer 1

Do you know its homo characteristic equation?

Answer 2

i want to learn about these type of equations,what do you mean homo characteristic, Chloro ?

Answer 3

when you see the right side = 0, It's homogeneous type!

Answer 4

how then is it solved ?

Answer 5

Solve for its root as second degree polynomial

Answer 6

is not this a differential equation of the second order ?

Answer 7

second order DE => solve by second degree polynomial!

Answer 8

can you show me how it is done ?

Answer 9

r^2 - 4r + 13 = 0

Answer 10

^this is called the "characteristic polynomial" of the DE

Answer 11

@KANNYTE can you solve it?

Answer 12

are'n t we going to form a differential equation first?

Answer 13

The DE goal is find y !

Answer 14

i don't understand ,can you please explain ?

Answer 15

Can you find the root for this:
r^2 - 4r + 13 = 0

Answer 16

yes i can

Answer 17

@KANNYTE are you really wanting to know why solving this quadratic will help us solve our problem?
are you just wanting to know why? because chlorophyll has provided the correct method

Answer 18

Pls show your roots here!

Answer 19

yes i want to know why

Answer 20

@KANNYTE follow the instruction, then:
Can you show the root for this:
r^2 - 4r + 13 = 0

Answer 21

we make an assumption that the solution will be some linear combination of\[y=e^{rt}\]it then follows that \[y'=re^{rt}\]and\[y''=r^2e^{rt}\]so under our assumption our eqn becomes\[r^2e^{rt}-4re^{rt}+13=0\]since \(e^{rt}\) can never be zero, that means the polynomial \[r^2-4r+13=0\]hence, if we find r, we know what our solution will look like.

Answer 22

r=2+6i and r=2-6i

Answer 23

Should you resolve it?

Answer 24

getting complex roots means we have to use Euler's formula:\[e^{it}=\cos t+i\sin t\]using some trig identities this means that when our solutions to the characteristic polynomial are a complex conjugate of the form\[\lambda\pm \mu i\]we know that our solution will be of the form\[y=c_1e^{\lambda}\cos(\mu t)+c_2e^{\lambda t}sin(\mu t)\]

Answer 25

if you want to see exactly how we get that formula read this link:
http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

Answer 26

typo above
the expected solution will have the form\[y(t)=c_1e^{\lambda t}\cos(\mu t)+c_2e^{\lambda t}\sin(\mu t)\]

Answer 27

so what will be the final solution of the equation?

Answer 28

Then, use initial values f(0) and f'(0( => C1 and C2

Answer 29

please don't just ask for the final answer, I'm not going to let it be that easy for you.
If that is what you want learn to use wolfram alpha.
chlorophyll and I have given you more than enough info to solve this problem I believe.

Answer 30

note: wolfram alpha is not the best when it comes to DE's
fair warning

Answer 31

@TuringTest Thanks for fulfill the asker's inquirement =)

Answer 32

likewise :)

Answer 33

i need your help so that i can solve this, i plead you to be patient with me guys,what you have been telling me earlier is like greek to me,can you explain from the basics of differential equations ? pliz

Answer 34

do you know what an integrating factor is?
can you solve a linear (first order) DE ?

Answer 35

yes i can

Answer 36

I think your roots are wrong; they should be\[r=2\pm3i=\lambda\pm\mu i\]

Answer 37

I told you that characterisitic polynomials of this form have a general solution\[y(t)=c_1e^{\lambda t}\cos(\mu t)+c_1e^{\lambda t}\sin(\mu t)\]plug in the values for \(\lambda\) and \(\mu\) that you found

after that, find plug in t=0 to find y(0) and y'(0), which will give you a system of equation which can be solved for c1 and c2, just as in the linear cases

Answer 38

I have to go, I hope I helped a bit
check out that link I gave you for further explanation

good luck!