Question #4886a

1 Answer
Jan 24, 2018

See below.

Explanation:

y=pcos(2x)+qsin(2x)

First differentiate:

dydx=2psin(2x)+2qcos(2x)

d2ydx2=4pcos(2x)4qsin(2x)

Now substitute into:

d2ydx2+4y

=4psin(2x)4qcos(2x)+4(psin(2x)+qcos(2x))

=4psin(2x)4qcos(2x)+4psin(2x)+4qcos(2x)

=4psin(2x)+4psin(2x)4qcos(2x)+4qcos(2x)

=0