Here are two equations:
eq1 = D[z[x], {x, 4}] + z[x] == 0;
eq2 = D[z[x], {x, 4}] + q^4*z[x] == 0;
They only differ from one another by the scale factor q^4, where q>0. I need to solve them in real numbers. If I solve the first one I get
DSolveValue[eq1, z[x], x]
(* E^(x/Sqrt[2]) C[1] Cos[x/Sqrt[2]] + E^(-(x/Sqrt[2])) C[2] Cos[x/Sqrt[2]] + E^(-(x/Sqrt[2])) C[3] Sin[x/Sqrt[2]] + E^(x/Sqrt[2]) C[4] Sin[x/Sqrt[2]] *)
which is convenient to look at. However, if I solve the second one, I get:
DSolveValue[eq2, z[x], x]
(* E^((-1)^(3/4) q x) C[1] + E^(-(-1)^(1/4) q x) C[2] + E^(-(-1)^(3/4) q x) C[3] + E^((-1)^(1/4) q x) C[4] *)
which is alredy less convenient. Since q is real and positive, it is obvious, that the solution is like the solution of eq1, in which we make a replacement: x->q*x
:
DSolveValue[eq1, z[x], x] /. x -> q*x
(* E^((q x)/Sqrt[2]) C[1] Cos[(q x)/Sqrt[2]] + E^(-((q x)/Sqrt[2])) C[2] Cos[(q x)/Sqrt[2]] + E^(-((q x)/Sqrt[2])) C[3] Sin[(q x)/Sqrt[2]] + E^((q x)/Sqrt[2]) C[4] Sin[(q x)/Sqrt[2]] *)
However, I cannot find a regular operation which would transform the solution of eq2 into this form.
Any idea?
DSolveValue[D[z[x], {x, 4}] + EulerGamma^4*z[x] == 0, z, x] /. EulerGamma -> q
. (Any positive symbolic constant could have worked just as well.) $\endgroup$ – J. M.'s ennui♦ Oct 5 '16 at 11:43Pi
instead of theEulerGamma
. However, I am preparing this for students, and poor students will go crazy with such tricks. They need something more regular. $\endgroup$ – Alexei Boulbitch Oct 5 '16 at 12:33Assumptions
, but it does nothing useful in this case. $\endgroup$ – J. M.'s ennui♦ Oct 5 '16 at 12:34