y4y0 y0y01 y00y0SolutionLet yz So given differential equatio

y^4-y\"=0 y(0)=y\'(0)=1 y\"\'(0)=0=y\"\"(0)

Solution

Let, y\'\'=z

So given differential equation becomes:

z\'\'=z

General solution to this is:

z=ae^{x}+be^{-x}

y\'\'=ae^{x}+be^{-x}

Integrating gives:

y\'=ae^{x}-be^{-x}+c

y=ae^{x}+be^{-x}+cx+d

y\'\'\'=ae^{x}-be^{-x}

y\'\'\'(0)=a-b=0

y\'\'(0)=a+b=0

Hence, a=b=0

y(0)=1

Hence,

c+d=1

y\'(0)=1

Hence, c=1

Hence, d=0

y(x)=x