Question

find the orthogonal trajectories of x^2+y^2=cx​

Answer 1

We're asked to find the orthogonal trajectory of the family of curves,

$\longrightarrow x^2+y^2=cx\quad\quad\dots(1)$

Differentiating wrt x,

$\longrightarrow 2x+2y\,\dfrac{dy}{dx}=c$

Putting this value of c in (1),

$\longrightarrow x^2+y^2=\left(2x+2y\,\dfrac{dy}{dx}\right)x$

$\longrightarrow x^2+y^2=2x^2+2xy\,\dfrac{dy}{dx}$

$\longrightarrow x^2-y^2+2xy\,\dfrac{dy}{dx}=0$

This is the differential equation of the given family of curves.

To obtain its orthogonal trajectory, replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$ because we know that the product of gradients equals -1 if the curves are perpendicular to each other.

Then,

$\longrightarrow x^2-y^2+2xy\left(-\dfrac{dx}{dy}\right)=0$

$\longrightarrow x^2-2xy\,\dfrac{dx}{dy}=y^2$

Multiply each term by $f(y)$ which is a function in y only.

$\longrightarrow x^2f(y)-2xy\,\dfrac{dx}{dy}f(y)=y^2f(y)$

$\longrightarrow x^2f(y)+2x\,\dfrac{dx}{dy}\left(-y\,f(y)\right)=y^2f(y)$

Assume $\dfrac{d}{dy}\left[-y\,f(y)\right]=f(y)$ such that the equation becomes,

$\longrightarrow\dfrac{d}{dy}\left[-x^2y\,f(y)\right]=y^2f(y)\quad\quad\dots(2)$

Now,

$\displaystyle\longrightarrow\dfrac{d}{dy}\left[-y\,f(y)\right]=f(y)$

$\displaystyle\longrightarrow\dfrac{d}{dy}\left[y\,f(y)\right]=-f(y)$

$\displaystyle\longrightarrow f(y)+y\,f'(y)=-f(y)$

$\displaystyle\longrightarrow y\,f'(y)=-2f(y)$

$\displaystyle\longrightarrow\dfrac{f'(y)}{f(y)}=-\dfrac{2}{y}$

Multilying by dy and integrating,

$\displaystyle\longrightarrow\int\dfrac{f'(y)}{f(y)}\ dy=-2\int\dfrac{1}{y}\ dy$

$\displaystyle\longrightarrow\log|f(y)|=-2\log|y|+\log\left|C_1\right|$

$\displaystyle\longrightarrow f(y)=\dfrac{C_1}{y^2}$

Then (2) becomes,

$\longrightarrow\dfrac{d}{dy}\left[-\dfrac{C_1x^2}{y}\right]=C_1$

$\longrightarrow\dfrac{d}{dy}\left[\dfrac{x^2}{y}\right]=-1$

$\longrightarrow d\left[\dfrac{x^2}{y}\right]=-dy$

Integrating,

$\displaystyle\longrightarrow \dfrac{x^2}{y}=-\int dy$

$\displaystyle\longrightarrow \dfrac{x^2}{y}=-y+c'$

$\displaystyle\longrightarrow\underline{\underline{x^2+y^2=c'y}}$

This is the orthogonal trajectory where c' is an arbitrary constant.