Question

the general solution of (D^2-4)y=0 is ​

Answer 1

Step-by-step explanation:

Assume Dn=dndxnDn=dndxn is the differential operator. Then, this is a fourth order linear differential equation.

(D4+4)y=0(D4+4)y=0

d4ydx4+4y=0d4ydx4+4y=0

The characteristic polynomial is fourth degree. It is not easy to compute the 4 roots of this equation, and the roots have both real and imaginary parts.

y4+4=0y4+4=0

(y2−2y+2)(y2+2y+2)=0(y2−2y+2)(y2+2y+2)=0

y1=(−1)−i,y2=(−1)+i,y3=1−i,y4=1+iy1=(−1)−i,y2=(−1)+i,y3=1−i,y4=1+i

The solutions are combinations of ex(cos(x)+exsin(x))ex(cos⁡(x)+exsin⁡(x)).

y(x)=Ae−xcos(x)+Be−xsin(x)+Cexcos(x)+Dexsin(x)y(x)=Ae−xcos⁡(x)+Be−xsin⁡(x)+Cexcos⁡(x)+Dexsin⁡(x)