Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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)