Solving differential equations with Wolfram Mathematica

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.

• Just because two mathematical expressions appear different, this does not mean they do not represent the same thing. Mar 25, 2015 at 8:06
• 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. Mar 25, 2015 at 8:50

Since you do not have initial conditions, the constants generated by Mathematica is solved for by comparing terms with $\sin$ and $\cos$ to make them match the book result. Then 4 equations solved to find the mapping

Clear[w, x, lam]
sol = w[x] /. First@DSolve[w''''[x] + 4*lam^4*w[x] == 0, w[x], x];
sol = ComplexExpand@sol

Comparing the above to book solution

book = Exp[lam x] (c[1] Cos[lam x] + c[2] Sin[lam x]) +
Exp[-lam x] (c[3] Cos[lam x] + c[4] Sin[lam x])

4 equations are solved

eq1 = C[1] + C[2] == c[3];
eq2 = C[1] - C[2] == -I c[4];
eq3 = C[3] + C[4] == c[1];
eq4 = C[4] - C[3] == -c[2] I;
map = Solve[{eq1, eq2, eq3, eq4}, {C[1], C[2], C[3], C[4]}]

So the above is the mapping between Mathematica solution and book solution. Verifying

Simplify[book - (sol /. map)]

• Oh, i get it, the imaginary number is STILL a number. Mar 25, 2015 at 8:53
• ComplexExpand@Re@sol is perhaps a shorter way to the book solution. Of course, then someone new to complex numbers might want to know why the real (and imaginary) part of a solution to a diff. eqn. (with real coefficients) also satisfies the diff. eqn. Mar 25, 2015 at 10:25

They are the same but they are written differently. Note that $\omega(0)= c_1+c_2+c_3+c_4$ in Mathematica but $\omega(0)= c_1+c_3$ in the book. So these are different coeffcients. So use the following:

ComplexExpand[DSolveValue[ω''''[x] + 4*λ^4*ω[x] == 0, ω[x], x]]

to get something similar to what you want. But you still need to redefine the coefficients. Also, note that $(-1)^{0.25}\sqrt{2}=1 + i$ and $(-1)^{0.75}\sqrt{2}=-1 + i$ !

• but there are no imaginary number in the textbook, does it mean that the answer is wrong? Mar 25, 2015 at 8:43