Question

Find a general solution for the differential equation with x as the independent variable:

$y^{\left(4\right)}+4y\text{'}\text{'}+4y=0$

Short answer

The general solution for the differential equation with x as the independent variable is

$y\left(x\right)=c_1\cos \left(\sqrt{2}x\right)+c_{2}x\cos \left(\sqrt{2}x\right)+c_3\sin \left(\sqrt{2}x\right)+c_{4}x\sin \left(\sqrt{2}x\right)$

Step-by-step solution

Auxiliary equation:

The auxiliary equation in this problem is $r^4+4r^2+4=0$. This can be factored as ${(r^2+2)}^2$=0. Therefore this equation has roots $r=\sqrt{2}i,-\sqrt{2}i,\sqrt{2}i,-\sqrt{2}i$, which we see are repeated and complex.

General solution:

The general solution to the given equation is given by

$y\left(x\right)=c_1\cos \left(\sqrt{2}x\right)+c_{2}x\cos \left(\sqrt{2}x\right)+c_3\sin \left(\sqrt{2}x\right)+c_{4}x\sin \left(\sqrt{2}x\right)$

Hence the final solution is $y\left(x\right)=c_1\cos \left(\sqrt{2}x\right)+c_{2}x\cos \left(\sqrt{2}x\right)+c_3\sin \left(\sqrt{2}x\right)+c_{4}x\sin \left(\sqrt{2}x\right)$