I have three coupled second order ODE's given as below
$x''[t] = -c_1*y'[t]-c_2*z[t]-c_3$
$y''[t] = -c_4*x'[t]$
$z''[t] = \frac{c_5}{c_6}*x[t]-c_6$
where $c_i$'s are know constants. The boundary conditions are
$x[-1]=x[1]=y[-1]=y[1]=z'[-1]=z'[1]=0$.
I followed the example given here How do I solve coupled ordinary differential equations?
{x, y, z} = {x, y, z} /.
Dsolve[{x''[t] == -c1*y'[t] - c2*z[t] - c3, y''[t] == -c4*x'[t],
z''[t] == -c5 + c5*x[t]/c6, x[-1] == 0, x[1] == 0, y[-1] == 0,
y[1] == 0, z'[-1] == 0, z'[1] == 0}, {x, y, z}, t] //
FullySimplify // First
I get the following output
ReplaceAll::reps: {Dsolve[{x''[t] == -c3 - c2 z[t] - c1 y'[t], <<8>>}, {x, y, z}, t]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing.
Set::shape: Lists {x, y, z} and {x, y, z} /. Dsolve[{x''[t] == -c3 - c2 z[t] - c1 y'[t], <<8>>}, {x, y, z}, t] are not the same shape.
Could someone please tell me what is wrong with my approach and please suggest me how to solve this system.
Thank you for your time
DSolve
instead ofDsolve
. $\endgroup${x, y, z}
should be{x[t], y[t], z[t]}
and I'd be inclined to applyFirst
beforeFullSimplify
. $\endgroup$