Find the family of curves c such that the tangent at an…

Find the family of curves $c$ such that the tangent at an arbitrary point cuts off on the $y$-axis (or $x$-axis) an intercept: (1) proportional to the ordinate (or abscissa) of the point of tangency; (2) proportional to the third power of the ordinate (or abscissa) of the point of tangency.

Case 1a: $y$-intercept $\propto y$

$y-xy'=ky\Rightarrow y'=\dfrac{(1-k)y}{x}$, separable: $\dfrac{dy}{y}=(1-k)\dfrac{dx}{x}\Rightarrow \boxed{y=Cx^{1-k}}$.

Case 1b: $x$-intercept $\propto x$

$x-\dfrac{y}{y'}=kx\Rightarrow y'=\dfrac{y}{(1-k)x}$, separable: $\boxed{y=Cx^{1/(1-k)}}$.

Case 2a: $y$-intercept $\propto y^3$

$y-xy'=ky^3\Rightarrow y'-\dfrac{y}{x}=-\dfrac{k}{x}y^3$ (Bernoulli, $n=3$). With $v=y^{-2}$: $v'+\dfrac{2v}{x}=\dfrac{2k}{x}$. Integrating factor $x^2$: $\dfrac{d}{dx}(x^2v)=2kx$, so $x^2v=kx^2+C$, i.e. $\dfrac{1}{y^2}=k+\dfrac{C}{x^2}$:

\[\boxed{\,y^2=\frac{x^2}{kx^2+C}.}\]

Case 2b: $x$-intercept $\propto x^3$

$x-\dfrac{y}{y'}=kx^3\Rightarrow y'=\dfrac{y}{x(1-kx^2)}$, separable. Partial fractions give $\ln|y|=\ln|x|-\tfrac12\ln|1-kx^2|+C_1$:

\[\boxed{\,y^2(1-kx^2)=C^2x^2.}\]