I want to solve this differential equation $$ u^{(4)}+a u^{(2)} +bu=0$$

I put $v=u^{(2)}.$ I obtain the new equation $v^{(2)}+a v+bu=0.$

What to do with $u?$ How to continue?


  • $\begingroup$ Are $a$ and $b$ constants? $\endgroup$ – Adrian Keister Aug 6 '19 at 14:50
  • $\begingroup$ yes @adrian keister $\endgroup$ – Vrouvrou Aug 6 '19 at 14:51
  • $\begingroup$ @complexmanifold: The notation $u^{(2)}$ means the second derivative of $u,$ not the square of $u.$ $\endgroup$ – Adrian Keister Aug 6 '19 at 15:00
  • $\begingroup$ @complexmanufold $u^{(2)}$ means the second derivative of u $\endgroup$ – Vrouvrou Aug 6 '19 at 15:01

I wouldn't do the substitution then, because it's unclear how to "not do" an operation like differentiation. I would go from the original DE, and substitute in the usual ansatz: $u=e^{\lambda x}$ (assuming $u=u(x).$) Then we obtain the quartic equation $\lambda^4+a\lambda^2+b=0.$ Here's where we would do the substitution $\alpha=\lambda^2,$ to obtain the quadratic $\alpha^2+a\alpha+b=0.$ The solution here is $$\alpha=\frac{-a\pm\sqrt{a^2-4b}}{2}. $$ We could get negative roots, making pairs of $\lambda$'s complex conjugates. We have $$\lambda=\pm\sqrt{\frac{-a\pm\sqrt{a^2-4b}}{2}}. $$ So the solution would be \begin{align*} u(x)= &A\exp\left(x\sqrt{\frac{-a+\sqrt{a^2-4b}}{2}}\right)+ B\exp\left(x\sqrt{\frac{-a-\sqrt{a^2-4b}}{2}}\right)+\\ &C\exp\left(-x\sqrt{\frac{-a+\sqrt{a^2-4b}}{2}}\right)+ D\exp\left(-x\sqrt{\frac{-a-\sqrt{a^2-4b}}{2}}\right) . \end{align*} You would need initial conditions to find $A, B, C,$ and $D.$ If there are complex solutions, you can translate to sine and cosine if you wish via $$e^{i\theta}=\cos(\theta)+i\sin(\theta). $$

  • $\begingroup$ thank you is there a methods when a and b are functions? $\endgroup$ – Vrouvrou Aug 6 '19 at 15:04
  • $\begingroup$ @Vrouvrou Not in general. $\endgroup$ – eyeballfrog Aug 6 '19 at 15:04
  • $\begingroup$ That's way harder! It would depend on the functions. Most often, if you need an analytical solution, series solutions are your only option. $\endgroup$ – Adrian Keister Aug 6 '19 at 15:04
  • 1
    $\begingroup$ "We're definitely going to get at least one negative root" Not if $a<0<b$. $\endgroup$ – J.G. Aug 6 '19 at 15:17
  • $\begingroup$ J.G. You're right, thanks! I'll edit. $\endgroup$ – Adrian Keister Aug 6 '19 at 15:27

Сharacteristic equation $t^4+at^2+b=0$. Then $t_{1,2,3,4}=\pm \sqrt{\frac{-a \pm \sqrt{a^2-4b}}{2}}$. Then $u=C_1e^{t_1x}+C_2e^{t_2x}+C_3e^{t_3x}+C_4e^{t_4x}$.

  • $\begingroup$ but $-a-\sqrt{a^2-4b}$ is negative ???? $\endgroup$ – Vrouvrou Aug 6 '19 at 20:07
  • $\begingroup$ If $a<0$ then positive. If $t_i$ is complex then there is Euler's formula. $\endgroup$ – Witold Aug 6 '19 at 21:30
  • $\begingroup$ what is euler formula? $\endgroup$ – Vrouvrou Aug 6 '19 at 22:17
  • $\begingroup$ Your problem is quite standard. Solution methods are described in textbooks ( math.psu.edu/tseng/class/Math251/Notes-HigherOrderLinEq.pdf ). $\endgroup$ – Witold Aug 6 '19 at 22:36

According to your idea the formulation is

$$ \begin{cases} u'' = v\\ v''=-a v-b u \end{cases} $$

You can now follow with the Laplace transform obtaining

$$ \begin{cases} s^2U(s) - su'(0)-u(0) = V(s)\\ s^2V(s)-s v'(0) - v(0) = -a V(s) - b U(s) \end{cases} $$


  • $\begingroup$ and if a and b are functions de can use this methods? $\endgroup$ – Vrouvrou Aug 6 '19 at 17:15
  • $\begingroup$ @Vrouvrou With $a(t),b(t)$ is not straightforward. $\endgroup$ – Cesareo Aug 6 '19 at 18:31
  • $\begingroup$ i don't understand what you mean? $\endgroup$ – Vrouvrou Aug 6 '19 at 18:32
  • $\begingroup$ @Vrouvrou For instance, if $a(t) = t^n$ the Laplace transform gives another differential equation now in $s$. For $a(t), b(t)$ Laplace transformable functions, after transforming we get an integral (convolutional) equation. Those diffrential or integral or integrodiffrential equations in $s$ can be quite cumbersome to solve. $\endgroup$ – Cesareo Aug 6 '19 at 18:36

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