how can i solve d4y/dt4 - λ4 y= 0
Since that is a linear equation with constant coefficients, you can immediately write down its "characteristic equation", [itex]x^4- \lambxa^4= 0[/itex]. That factors as [itex](x^2- \lambda^2)(x^2+ \lambda^2)= (x- \lambda)(x+ \lambda)(x- i\lambda)(x+ i\lambda)= 0[/itex] and so has characteristic values [itex]\lambda[/itex], [itex]-\lambda[/itex], [itex]i\lambda[/itex], and [itex]-i\lambda[/itex]. Do you know what to do with those?
hmm... i dont really understand. can u show me a clearer working? wat you mean by x^4- \lambxa^4= 0? thanks
What kind of explanation would you understand? Do you know anything at all about linear differential equations? I need to know what you do understand before I can explain much more.
oh how did you solve x^2 + λ^2?
I didn't solve [itex]x^2+ \lamba^2[/itex]. That is not an equation. I did solve [itex]x^2+ \lambda^2= 0[/itex] by the obvious method: I subtracted [itex]\lambda^2[/itex] from both sides to get [itex]x^2= -\lambda^2[/itex] and then took the square root of both sides. But now, in addition to my previous questions, which you still haven't answered, why are you trying to do differential equations if you don't know how to solve a simple quadratic equation?
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