Answer 1

y = e^ax. cos bx

`dy/dx=ae^(ax).cosbx-be^(ax).sinbx.........(i)`

`dy/dx=ay-be^(ax).sinbx`

`(d^2y)/(dx^2)=ady/dx-b(ae^(ax).sinbx+be^(ax).cosbx)`

`(d^2y)/(dx^2)=ady/dx-abe^(ax).sinbx-b^2e^(ax).cosbx`

`(d^2y)/(dx^2)=ady/dx-a(ay-dy/dx)-b^2y ` [Substituting be^ax sinbx from(i)]

`(d^2y)/(dx^2)=ady/dx-a^2y+ady/dx-b^2y`

`therefore (d^2y)/(dx^2)-2ady/dx+(a^2+b^2)y=0`

Hence Proved