Extraordinary Differential Equations #7

Calculus Level 3

The real-valued solution curve $y=y(x)$ to $\frac{\sqrt{3}}{2}y'=\frac{y^2+y+1}{\sqrt{1+4x^2}}$ has a $y$ -intercept at $(0,-\frac{1}{2})$ . Determine the area of the right-angled triangle in the fourth quadrant bounded by the origin, the $x$ -intercept and the $y$ -intercept of the solution curve.

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

\frac{1}{16} \cos{\frac{\pi}{3}}

\frac{1}{8} \sinh{\frac{\pi}{3}}

\frac{1}{2} \sin{\frac{\pi}{3}}

\frac{1}{2} \tan^2 {\frac{\pi}{3}}

\frac{1}{4} \tanh{\frac{\pi}{3}}

\frac{1}{4} \cosh{\frac{\pi}{3}}

1 solution

Wee Xian Bin
Feb 1, 2017

Relevant wiki: Derivatives of Inverse Hyperbolic Trigonometric Functions, and Integrals that Result in Hyperbolic Trigonometric Functions

This is a separable ODE: $\frac{\sqrt{3}}{y^2+y+1} dy =\frac{2}{\sqrt{1+4x^2}} dx$ $2 \tan^{-1} (\frac{2y+1}{\sqrt{3}}) = \sinh^{-1}{(2x)}+c$ Using the $y$ -intercept's coordinates as a constraint we get $c=0$ . Therefore, the $x$ -intercept is $(\frac{1}{2} \sinh{\frac{\pi}{3}}, 0)$ and the required triangle's area is $\frac{1}{8} \sinh{\frac{\pi}{3}}$ .

Extraordinary Differential Equations #7

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

1 solution

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