So i saw this differential equations in my textbook
$\frac{{{d^4}\omega }}{{d{x^4}}} + 4{\lambda ^4}\omega = 0$
and i figured why not solve it with majestic Wolfram Mathematica, so i write this code:
DSolve[ω''''[x] + 4*λ^4*ω[x] == 0, ω[x],x]
the result:
$\omega (x)\to c_1 e^{(-1)^{3/4} \sqrt{2} \lambda x}+c_2 e^{-\sqrt[4]{-1} \sqrt{2} \lambda x}+c_3 e^{-(-1)^{3/4} \sqrt{2} \lambda x}+c_4 e^{\sqrt[4]{-1} \sqrt{2} \lambda x}$
which is a lot different from what my textbook presented:
$\omega (x) = {e^{\lambda x}}({c_1}\cos \lambda x + {c_2}\sin \lambda x) + {e^{ - \lambda x}}({c_3}\cos \lambda x + {c_4}\sin \lambda x)$
where did i go wrong, how can i get the same answer?
unpacking the $\sqrt[4]{{ - 1}}$ term as follow:
for $\sqrt[4]{-1}=\frac{1+i}{\sqrt{2}}$ and $(-1)^{3/4}=\frac{i-1}{\sqrt{2}}$
$\begin{array}{c} \omega (x) = {c_1}{e^{( - 1 + i)\lambda x}} + {c_2}{e^{( - 1 - i)\lambda x}} + {c_3}{e^{(1 - i)\lambda x}} + {c_4}{e^{(1 + i)\lambda x}}\\ = {e^{ - \lambda x}}({c_1}{e^{i\lambda x}} + {c_2}{e^{ - i\lambda x}}) + {e^{\lambda x}}({c_3}{e^{ - i\lambda x}} + {c_4}{e^{i\lambda x}}) \end{array} $
Acorrding to Eulers formula: ${e^{ix}} = \cos x + i\sin x$
$\begin{array}{c} \omega (x) = {e^{ - \lambda x}}({c_1}(\cos \lambda x + i\sin \lambda x) + {c_2}(\cos \lambda x - i\sin \lambda x)) + {e^{\lambda x}}({c_3}(\cos \lambda x - i\sin \lambda x) + {c_4}(\cos \lambda x + i\sin \lambda x))\\ = {e^{ - \lambda x}}(({c_1} + {c_2})\cos \lambda x + ({c_1} - {c_2})i\sin \lambda x) + {e^{\lambda x}}(({c_3} + {c_4})\cos \lambda x + ({c_4} - {c_3})i\sin \lambda x) \end{array} $
Simplify:
$\omega (x) = {e^{ - \lambda x}}({C_1}\cos \lambda x + {C_2}i\sin \lambda x) + {e^{\lambda x}}({C_3}\cos \lambda x + {C_4}i\sin \lambda x)$
There is an extra i comparing to the textbook.
There is an extra i comparing to the textbook
it is normal to rename the whole thing to a constant. i.e. $(c_1-c_2)i$ is renamed to new constant, say $C[1]$. So the $i$ goes with the constant and is not left out. This is standard way of reformulating solutions to ode's so the solution is written in terms of trig functions. $\endgroup$ – Nasser Mar 25 '15 at 8:50