The Student Room Group
Reply 1
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
Reply 2
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?
Reply 3
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
Reply 4
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!