Answer to Question #242887 in Differential Equations for Phyroehan

Question

Equation of Rectangular Curves.

1. For a certain curve, the point of contract of each tangent to its bisects the part of the tangent terminating on the coordinate axes. Find the equation of the curve.

2. The area bounded by the curve, the x axis, a fixed ordinate and a variable ordinate is proportional to the difference between the ordinates. Find the equation of the curve.

Answer 1

1.

The equation of tangent to the curve at $(x,y)$ is

$Y-y=\frac{dy}{dx}(X-x)$

Put $Y=0,$ then $A=(x-y\frac{dy}{dx},0)$

and $X=0,$ then $B=(0,y-x\frac{dy}{dx})$

It is given that,

$\frac{x-y\frac{dy}{dx}}{2}=x$ and $\frac{y-x\frac{dy}{dx}}{2}=y$

$\implies x-y\frac{dy}{dx}=2x$ and $y-x\frac{dy}{dx}=2y$

$\implies y\frac{dy}{dx}=-x$ and $x\frac{dy}{dx}=-y$

$\implies \frac{dx}{x}+\frac{dy}{y}=0$

Integrate both sides

$\ln|x|+\ln|y|=\ln|C|\\ \implies xy=C$

This is the required equation of the curve.

2.

Let $y=f(x)$ be the equation of the curve, $x_0$ be the abscissa of the point with the fixed ordinate, $x$ be the abscissa of the point with the variable ordinate. According to the conditions we have an equation:

$\int_{x_0}^x f(t)dt=C(f(x)-f(x_0)),$

where $C$ is the constant of proportionality. Let's differentiate this equation with respect to $x$ and we will have differential equation

$f(x)=Cf'(x).$

Solution of this differential equation is

$f(x)=C_1e^{(\frac{x}{C})}$ , where $C_1$ is an arbitrary real constant.

So we have a set of curves which satisfy the equation

$y=C_1e^{(\frac{x}{C})}$