How do you find #(d^2y)/(dx^2)# given #y+siny=x#?

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Answer 1 · 2016-08-02T17:01:24

#sin y/(1+cos y)^3#

Here, x=y+sin y.

So, 1=y'+y'cos y=(1+cos y)y'.

And so, y'=1/(1+cos y)

Now, #y''=-1/(1+cos y)^2(- sin y) y'#

#=sin y/(1+cos y)^3#

Answer 2 · 2016-08-02T17:12:00

#y'' = (sin(y))/(1+cos(y))^3#

Given #y+sin(y)-x=0# there is a functional dependency between #x# and #y#. We will rewrite the former equation as:

#f(x,y)=y(x)+sin(y(x))-x=0#

We know

#df(x,y)=f_x dx+f_y dy = 0# and

#(dy)/(dx)=-f_x/(f_y)=y'(x) = 1/(1 + cos(y(x)))#

so we have now the functional relationship

#y'(x) -1/(1 + cos(y(x)))=0#

deriving regarding #x# we obtained

#y''(x)-(sin(y(x))y'(x))/(1+cos(y(x)))^2=0#

substituting for #y'(x)# we get

#y''(x) = (sin(y(x)))/(1+cos(y(x)))^3# or simply

#y'' = (sin(y))/(1+cos(y))^3#

Attached the plot of #y(x)# (blue) and #y''(x)# (red)

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