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    Does anyone know how to solve this system of equations:
    y''=z
    z''=y
    with boundary conditions y(0)=0, y(π/2)=1, z(0)=0, z(π/2)=1,

    where y'' means d²y/dx² ?
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    (Original post by m277)
    Does anyone know how to solve this system of equations:
    y''=z
    z''=y
    with boundary conditions y(0)=0, y(π/2)=1, z(0)=0, z(π/2)=1,

    where y'' means d²y/dx² ?
    One way is to get y'''' = y which has general solution

    y = Asinx + Bcosx + Cexp(x) + Dexp(-x)

    and then

    z = -Asinx - Bcosx + Cexp(x) + Dexp(-x)

    Your boundary conditions then give you four equations in A,B,C,D
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    (Original post by RichE)
    One way is to get y'''' = y which has general solution

    y = Asinx + Bcosx + Cexp(x) + Dexp(-x)

    and then

    z = -Asinx - Bcosx + Cexp(x) + Dexp(-x)

    Your boundary conditions then give you four equations in A,B,C,D
    Thanks for this. I considered y''''=y but I didn't know what the general solution was.

    How did you come up with this general solution? ie where did it come from?
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    (Original post by m277)
    Thanks for this. I considered y''''=y but I didn't know what the general solution was.

    How did you come up with this general solution? ie where did it come from?
    From the auxiliary equation m^4 = 1
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    (Original post by RichE)
    From the auxiliary equation m^4 = 1
    ahhh, I see.

    You factorise so that (m²-1)(m²+1)=0
    and then use the auxilliary equation that I already know. So you end up having a linear combination of the complementary functions to m²-1=0 and m²+1=0.

    Thank you. My tutor won't rip me to shreds now!
 
 
 
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