# Solving a fourth order differential equation

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?

Thanks!

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

## 3 Answers

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 $$e^{i\theta}=\cos(\theta)+i\sin(\theta).$$

• thank you is there a methods when a and b are functions? – Vrouvrou Aug 6 '19 at 15:04
• @Vrouvrou Not in general. – eyeballfrog Aug 6 '19 at 15:04
• That's way harder! It would depend on the functions. Most often, if you need an analytical solution, series solutions are your only option. – Adrian Keister Aug 6 '19 at 15:04
• "We're definitely going to get at least one negative root" Not if $a<0<b$. – J.G. Aug 6 '19 at 15:17
• J.G. You're right, thanks! I'll edit. – 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 $$u=C_1e^{t_1x}+C_2e^{t_2x}+C_3e^{t_3x}+C_4e^{t_4x}$$.

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

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

etc.

• and if a and b are functions de can use this methods? – Vrouvrou Aug 6 '19 at 17:15
• @Vrouvrou With $a(t),b(t)$ is not straightforward. – Cesareo Aug 6 '19 at 18:31
• i don't understand what you mean? – Vrouvrou Aug 6 '19 at 18:32
• @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. – Cesareo Aug 6 '19 at 18:36