Extraordinary Differential Equations #9

Calculus Level 3

Derive an expression of the arc length of the curve $y=y(x)$ governed by the differential equation $\frac{d^4y}{dx^4}=-2 \text{ sech}^2 (x) \tanh(x)$ and the initial conditions $y'(0)=0, y''(0)=0, y'''(0)=1$ between the points $(a,y(a))$ and $(b,y(b))$ where $b>a$ are real numbers.

(This problem is part of the set Extraordinary Differential Equations .)

\int_a^b \sqrt{1+\ln{(\cosh{x})}} \mathrm{d}x

\int_a^b \tanh{x} \mathrm{d}x

\int_a^b \ln{(\cosh{x})} \mathrm{d}x

\int_a^b \sqrt{1+\text{sech}^4 {x}} \mathrm{d}x

\int_a^b \sinh{x} \mathrm{d}x

\int_a^b \tanh{x} \text{ sech }x \mathrm{d}x

\int_a^b \sqrt{1+(\ln{(\cosh{x})})^2} \mathrm{d}x

\int_a^b \sqrt{1+\tanh^2{x}} \mathrm{d}x

1 solution

Wee Xian Bin
Feb 4, 2017

If $\frac{d^4y}{dx^4}=-2 \text{ sech}^2 (x) \tanh(x)$ then $\frac{d^3y}{dx^3}=\text{ sech}^2 (x)+c_3 \overset{y'''(0)=1}{\implies} \frac{d^3y}{dx^3}=\text{ sech}^2 (x)$ $\frac{d^2y}{dx^2}=\tanh{x}+c_2 \overset{y''(0)=0}{\implies} \frac{d^2y}{dx^2}=\tanh{x}$ $\frac{dy}{dx}=\ln{(\cosh{x})}+c_1 \overset{y'(0)=0}{\implies} \frac{dy}{dx}=\ln{(\cosh{x})}$ The arc length is $\int_a^b \sqrt{1+(\frac{dy}{dx})^2} \mathrm{d}x= \int_a^b \sqrt{1+(\ln{(\cosh{x})})^2} \mathrm{d}x$

Extraordinary Differential Equations #9

1 solution

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