Question

Show that the differential equation (on the left) is a solution of the function (on the right)

d^2u/dt^2 = a^2 * (d^2u/dx^2) u(x,t) = f(x-at) + g(x+at)

Answer 1

We have to show that

`(\partial ^(2)u)/(\partial t^(2))=a^(2)(\partial ^(2)u)/(\partial x^(2))`

for
`(\partial ^(2)u)/(\partial t^(2))` we have

`(\partial ^(2)u)/(\partial t^(2))=a^(2)(\partial ^(2)u)/(\partial x^(2))`

`(\partial ^(2)u)/(\partial t^(2))=(\partial ^(2)[f(x-at)+g(x+at)])/(\partial t^(2))`

`=(\partial )/(\partial t)[(\partial[f(x-at)+g(x+at)] )/(\partial t)]`

`(\partial )/(\partial t)[-a\cdot f'(x-at)+a\cdot g'(x+at)]`

`=a^(2)f''(x-at)+a^(2)g''(x+at)`

`=a^(2)[f''(x-at)+g''(x+at)].............(i)`

similarly,

`(\partial ^(2)u)/(\partial x^(2))=(\partial ^(2)[f(x-at)+g(x+at)])/(\partial x^(2))`

`=(\partial )/(\partial x)[(\partial[f(x-at)+g(x+at)] )/(\partial x)]`

`=(\partial )/(\partial x)[f'(x-at)+g'(x+at)]`

`=f''(x-at)+g''(x+at).......(ii)`

Comparing i and ii we get

`a^(2)(\partial ^(2)u)/(\partial x^(2))=(\partial ^(2)u)/(\partial t^(2))`

Hence proved