4th order linear ordinary differential equation

While I was solving an integral using Feynman Integration, I came across the following differential equation:

$$y’’’’-y’’+y=0$$

I tried substituting $y$ with an exponential function which failed. Can someone else show me how to solve it?

Solution 1:

$$λ^4−λ^2+1$$ can be completed to a square by variation of the middle term, making it more negative in the process $$=(λ^2+1)^2−3λ^2=(λ^2+\sqrt3λ+1)(λ^2-\sqrt3λ+1)$$ Now one can apply the usual solution formulas for quadratic equations (with real coefficients but complex roots).

Solution 2:

Hint

The characteristic equation of you ODE is $x^4-x^2+1=0$ which, when multiplied in $x^2+1$, yields $x^6+1=0$. Hence, the roots of $x^4-x^2+1$ are those of $x^6=-1$ except those of $x^2=-1$.