How do you determine #(d^2y)/(dx^2)# given #sinx-4cosy=y#?

Question

1309 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-07T00:15:26

Differentiate with respect to variable x the given equation

#(sinx-4cosy)'=(y)'#

#cosx+4siny*y'=y'#

Differentiate again to get

#-sinx+4cosy*y' * y'+4siny*(y' ')=y' '#

#-sinx+4cosy*(y')^2=y' '*(1-4siny)#

#y' '=[-sinx+4*cosy*(y')^2]/[1-4siny]#

Note : We suppose that y is a funtion of x hence #y=y(x)#