# Solving the differential equation $y^{(4)} + 4y = 0$

I have trouble solving the differential equation $y^{(4)} + 4y = 0$

This is a linear differential equation with constant coefficients.

Solving for the roots of the associated polynomial gives:

$$r^4 + 4r = 0 \iff r(r^3 + 4) = 0 \iff r(r+\sqrt[3]{4})(r^2 - \sqrt[3]{4}r + \sqrt[3]{4}^2) = 0$$

I get very ugly solutions for the roots of this quadratic equation. My book says the answer should be $$y = e^x(c_1\cos(x) + c_2\sin(x))+ e^{-x}(c_3\cos(x) + c_4\sin(x))$$ and wolfram alpha confirms this.

But, in general, if $a + bi$ is a complex root of the polynomial of multiplicity $1$ then $e^{ax}(\cos(bx) + \sin(bx))$ will be a solution, so it seems I won't be able to reach that solution.

Any help will be greatly appreciated!

• Your characteristic equation should be $r^4 + 4=0$. – Zain Patel May 4 '17 at 19:52

## 2 Answers

In general, the characteristic equation of the linear DE $a_n y^{(n)} + \cdots a_0y = 0$ is $a_n r^n + \cdots + a_0 = 0$.

In your case, this means your equation should be $r^4 + 4 = 0.$ This gives $r = (-1)^{1/4}\sqrt{2}$. That is $r = \sqrt{2}e^{\pm i \pi/4}$ and $r = \sqrt{2}e^{\pm 3i\pi/4}$.

$$r^4+4=(r^2+2r+2)(r^2-2r+2)$$

Then roots of equation $r^4+4=0\;$ is $\;\pm1\pm i$ and solution of ODE is $$y = e^x(c_1\cos(x) + c_2\sin(x))+ e^{-x}(c_3\cos(x) + c_4\sin(x))$$

• Thanks, though the question was already solved and the diff course is already done haha. – user370967 Jun 4 '18 at 20:59
• Yes, but a year later I'm struggle with the same diff equation... – Ron Jensen Jun 11 at 3:05