Extraordinary Differential Equations #8

Calculus Level 3

Orthogonal trajectories are a family of curves (say $F_2$ ) in a plane that intersect a given family of curves (say $F_1$ ) at right angles.

Suppose $F_1$ is given by $y=c e^{-x^2}$ where $y=y(x)$ and $c$ is an arbitrary constant. Derive an equation for $F_2$ where $c$ is an arbitrary constant of integration.

Hint: If $F_1$ is defined by a differential equation of the form $y'=f(x,y)$ , then $F_2$ is defined by the differential equation $y'=-\frac{1}{f(x,y)}$ ( why? ).

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

y=c e^{x^{-2}}

y^2=\ln{(\pm x)}+c

y=\ln{(\pm x)}+c

y=c e^{-x^2}

y=c e^{x^2}

x=c e^{-y^2}

y^2=c-\ln{(\pm x)}

1 solution

Wee Xian Bin
Feb 1, 2017

Relevant wiki: First Order Linear Differential Equations

First we derive the ODE for $F_1$ by differentiate both sides of the given equation: $y'=(-2x)(ce^{-x^2}) \implies y'=-2xy$ Then the ODE for $F_2$ is $y'=\frac{1}{2xy}$ $\int 2y dy = \int \frac{1}{x} dx$ the solution to which is $y^2=\ln{(\pm x)}+c$

Extraordinary Differential Equations #8

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

1 solution

0 pending reports