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# is it i=-1 or -i=1 (or am i wrong on both counts?) watch

1. i've forgotten what that rule is. has it got a name?

because i've got a square root containing -4, [equation being 2^2 - 4(1)(2)], and i know i have to put sqrt4i or something along those lines. any help??

2. Well, both are wrong.
But i^2=-1.
3. and . i is one of the square roots of -1.
4. What?

(if taking the 'positive' root), and , so your answer is pretty much straightforward from there.

But if your equation is , then that's different to or whatever; is only for if the minus is inside the root.

EDIT: Beaten to it three times!
5. yes! atleast i was close! thanks
6. ok sorry confused again! :P

y'' + 2' + 2 = 0

(-2 (+-) sqrt(2^2 - 4(1)(2)))/2(1)

= (-2 (+-) sqrt(-4))/2

i remember the teacher doing something like:

(-2 (+-) sqrt(4i))/2 or similar, is that right?
7. (Original post by furryvision)
ok sorry confused again! :P

y'' + 2' + 2 = 0

(-2 (+-) sqrt(2^2 - 4(1)(2)))/2(1)

= (-2 (+-) sqrt(-4))/2

i remember the teacher doing something like:

(-2 (+-) sqrt(4i))/2 or similar, is that right?
See nuodai's post. Informally, sqrt(-4) = sqrt(4)*sqrt(-1) = sqrt(4) * i. Definitely not sqrt(4i) (because there you've just replaced -1 with i, which we told you was wrong ).
8. No, because the should be outside the root. .

But, by y'' do you mean y² or ? Because if you mean , you're going about this very much the wrong way, although you would need to solve the equation to get the auxiliary equation so the bit isn't completely wasted.
9. (Original post by nuodai)
you would need to solve the equation to get the auxiliary equation.
That's what I assumed she was doing.
10. (Original post by nuodai)
No, because the should be outside the root. .

But, by y'' do you mean y² or ? Because if you mean , you're going about this very much the wrong way, although you would need to solve the equation to get the auxiliary equation so the bit isn't completely wasted.
thanks, i'm making sense of it all now! y" is the d^2y/dx^2, but it equals cos(x), not 0, i just said that to make the question easier. it says in my notes to find the general solutions first and my teacher got the 1+-i so i understand now. thanks!
11. (Original post by furryvision)
thanks, i'm making sense of it all now! y" is the d^2y/dx^2, but it equals cos(x), not 0, i just said that to make the question easier. it says in my notes to find the general solutions first and my teacher got the 1+-i so i understand now. thanks!
Fair enough; in such a case you don't need to do any quadratic equation. To get your solution, you need to:
1. Let , and solve for A and B by substituting in and differentiating etc. to give your particular integral
2. Let (ignore the cos x here) and solve for giving two solutions and , so your complementary function is
3. Sum the complementary function and particular integral

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