Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

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 1: 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.

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