Question #4886a

Question

871 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-24T22:25:05

See below.

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

First differentiate:

#dy/dx=-2p sin(2x)+2qcos(2x)#

#(d^2y)/(dx^2)=-4p cos(2x)-4qsin(2x)#

Now substitute into:

#(d^2y)/(dx^2)+4y#

#=-4p sin(2x)-4qcos(2x)+4(p sin(2x)+qcos(2x))#

#=-4p sin(2x)-4qcos(2x)+4p sin(2x)+4qcos(2x)#

#=-4p sin(2x)+4p sin(2x)-4qcos(2x)+4qcos(2x)#

#=0#