Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    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??
    Offline

    19
    i^2 = -1

    Offline

    14
    ReputationRep:
    Well, both are wrong.
    But i^2=-1.
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    i^2 = -1 and (-i)^2 = -1. i is one of the square roots of -1.
    • PS Helper
    Offline

    14
    PS Helper
    What?

    i^2 = -1 \Rightarrow i = \sqrt{-1} (if taking the 'positive' root), and \sqrt{-4} = \sqrt{-1}\sqrt{4}, so your answer is pretty much straightforward from there.

    But if your equation is 2^2 - \sqrt{4}, then that's different to 2^2 - \sqrt{-4} or whatever; i is only for if the minus is inside the root.

    EDIT: Beaten to it three times!
    • Thread Starter
    Offline

    2
    ReputationRep:
    yes! atleast i was close! thanks
    • Thread Starter
    Offline

    2
    ReputationRep:
    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?
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (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 :p:).
    • PS Helper
    Offline

    14
    PS Helper
    No, because the i should be outside the root. \dfrac{-2 \pm \sqrt{4}i}{2} = \dfrac{-2 \pm 2i}{2} = 1 \pm i.

    But, by y'' do you mean y² or \dfrac{\mbox{d}^2y}{\mbox{d}x^2}? Because if you mean \dfrac{\mbox{d}^2y}{\mbox{d}x^2}, you're going about this very much the wrong way, although you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation so the 1 \pm i bit isn't completely wasted.
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by nuodai)
    you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation.
    That's what I assumed she was doing.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by nuodai)
    No, because the i should be outside the root. \dfrac{-2 \pm \sqrt{4}i}{2} = \dfrac{-2 \pm 2i}{2} = 1 \pm i.

    But, by y'' do you mean y² or \dfrac{\mbox{d}^2y}{\mbox{d}x^2}? Because if you mean \dfrac{\mbox{d}^2y}{\mbox{d}x^2}, you're going about this very much the wrong way, although you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation so the 1 \pm i 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!
    • PS Helper
    Offline

    14
    PS Helper
    (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 y = A\sin x + B\cos x, and solve for A and B by substituting in and differentiating etc. to give your particular integral
    2. Let \lambda^2 + 2\lambda + 2 = 0 (ignore the cos x here) and solve for \lambda giving two solutions \lambda_1 and \lambda_2, so your complementary function is Ce^{\lambda_1} + De^{\lambda_2}
    3. Sum the complementary function and particular integral
 
 
 

3,463

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.