Question

If y = e^ax. cos bx, then prove that

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

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 sin bx 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