Question

y''-8y'+12y=0, y(0)=4, y(1)=2

Answer 1

So this is a homogenous differential equation. So what we want to do is take the characteristic polynomial and find the roots of that equation. Think you can do that part?

Answer 2

yes i have done that part y=c1e^(2x)+c2e^(6x)

Answer 3

Hmm....Did you change the equation or was I just not paying attention? O.o

Answer 4

sorry i made mistake it was 12y not 8y

Answer 5

no i changed it

Answer 6

Ah, okay : ) Then yes, your complimentary solution is correct then. So the first of your initial conditions gives you y(0) = 4. So this means when x is 0, I need to be equal to 4

4 = c1e^(2*0)+c2e^(6*0).

This of course simplifies to

4 = c1 + c2. So not enough information to solve for those coefficients. So we need to use the other initial condition. Did you mean that second condition to be y(1) or y'(1)?

Answer 7

yes i have this first question same but second question is not y' because the second value is given is y not y'

Answer 8

I just wanted to sure it was supposed to be y, you often see one condition with y and the other with y'

Okay, so using the 2nd condition y(1) = 2, we have

2 = c1e^2 + c2e^6. So this gives us two equations which we can use to solve for c1 and c2.

4 = c1 + c2
2 = c1e^2 + c2e^6

So now its just using whichever method of solving a system of equations you prefer.

Answer 9

i know we have two questions and i get c1=10.908 and c2= -14.908 but i am not sure if they are right

Answer 10

Well, let's check. So if I were to just do substitution, I'll do 4-c2 = c1

2 = (4-c2)e^2 + c2e^6
2 = 4e^2 - c2e^2 + c2e^6
2-4e^2 = c2(-e^2 + e^6)
\[c_{2}=\frac{ 2-4e^{2} }{ e^{6}-e^{2} }\approx -.07\] With that value, Ill plug it in for c2 into the first equation:

4 = c1 - .07
\(c_{1}\approx\ 4.07\)

Answer 11

the way i did i found the answer first for e^2 and e^6 for second question and put it back to first question

Answer 12

no when i put c1 as 4.07 and c2=-0.07 i didnt get right it

Answer 13

thank you

Answer 14

well, those are rounded off answers. More decimals to those answers would give:
\(c_{1} \approx4.06958\)
\(c_{2} \approx-.069579\)

Is there any requirement on rounding or if it wants an exact solution or not?

Answer 15

ok

Answer 16

Another possible approach is using the Laplace transform, if you're familiar with it.