Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Curve C: y = ln(1+ cos(x)), show that...?

d^(4)y/dx^(4) + (e^(-y))(dy/dx)^(2) + 2e^(-2y) = 0

So here's my working:..

dy/dx = -sin(x)/(cos(x)+1)

so (dy/dx)^(2) = -sin^(2)(x)/(1+cos(x))^(2)

d^(4)y/dx^(4) = (cos(x) -2)/(cos(x)+1)^(2)

e^(-y) = 1/(cos(x)+1)

e^(-2y) = 1/(cos(x)+1)^(2)

So, plugging back in...

(cos(x) -2)/(cos(x)+1)^(2) + [1/(cos(x)+1)][sin^(2)(x)/(1+cos(x))^(2)] + 2/(cos(x)+1)^(2)

= [(cos(x)-2)/(cos(x) +1)^(2)] + [(sin^(2)(x))/(1+cos(x))^(3)] + [2/(1+cos(x))^(2)]

simplifying to:

1/(cos(x) +1)^(2)

???

Update:

* just to add: by (dy/dx)^(2) I do mean the square of the first derivative, not the second derivative.

Relevance
• Curve C: y = ln(1+ cos(x)), show that...?

d^(4)y/dx^(4) + (e^(-y))(dy/dx)^(2) + 2e^(-2y) = 0

y'= -sinx/(1+cos x), y''= -cosx/(1+cos x)-sin^2x/(1+cos x)^2=-(cos x(1+cos x)+sin^2x)/(1+cos x)^2=-1/(1+cos x)=-e^-y, y'''= y' e^-y, y'''=-y'^2e^-y-y''e^-y=-y'^2e^-y+e^-2y. I get the answer  y''''+y'^2 e^-y-e^-2y=0. Check again.

• Looks like you went astray on the 2nd derivative.  I got:

y' = 1/(1 + cos x) * (-sin x) = - (sin x)/(1 + cos x)  . . . . same as you, then

y'' = - [(1 + cos x)(sin x)' - (sin x)(1 + cos x)'] / (1 + cos x)²

= - [(1 + cos x)(cos x) - (sin x)(-sin x)] / (1 + cos x)²

= - (cos x + cos² x + sin² x) / (1 + cos x)²

= - (1 + cos x) / (1 + cos x)²

= -1 / (1 + cos x)

Doesn't look equivalent to yours, but maybe I'm missing something.