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 wrote this code:

DSolve[[Omega]''''[x] + 4*[Lambda]^4*[Omega][x] == 0, [Omega][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 follows:

for $\sqrt[4]{-1}=\frac{1+i}{\sqrt{2}}$ and $\sqrt[4]{-1}=\frac{-1+i}{\sqrt{2}}$

$\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} $


$\omega (x) = {e^{ - \lambda x}}({C_1}\cos \lambda x + {C_2}i\sin \lambda x) + {e^{\lambda x}}({C_3}\cos \lambda x + {C_4}i\sin \lambda x)$

There is an extra i comparing to the textbook.

  • $\begingroup$ You need to unpack the $\sqrt[4]{-1}$ terms $\endgroup$ – Henry Mar 25 '15 at 7:08
  • $\begingroup$ Apply ComplexExpand to the solution Mathematica gives you, and you will find it in the form you want. By the way, this should perhaps be moved to the Mathematica SE? $\endgroup$ – mickep Mar 25 '15 at 8:27

It is the same answer, once in real and once in complex form. Note that the 4th roots of $-4λ^4$ are $\pmλ\pm i·λ$, with independent sign selection for 4 roots. One could guess that Mathematica understands $\sqrt[4]{-1}=(-1)^{1/4}=\frac{1+i}{\sqrt{2}}$ and $(-1)^{3/4}=\frac{-1+i}{\sqrt{2}}$.

The combination of complex conjugate roots gives the exponential-trigonometric form.


Let $C_5 = C_2 i$ and $C_6 = C_4 i$. This will also catch wholly real solutions for wholly imaginary $C_2$ and $C_4$


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