Answer to Question #95320 in Differential Equations for Pooja panghal

Question

Find the orthagonal trijectries of the family of parabolas x=cy^2 and sketch their graph

Answer 1

Differeniate $x=cy^2$: $dx=2cydy$ , so $y'=\frac{dy}{dx}=\frac{1}{2cy}$

Express $c$ from the initial equation: $c=\frac{x}{y^2}$. Then we have the differential equation of the family $x=cy^2$ :$y'=\frac{1}{2cy}=\frac{1}{2y}\cdot\frac{y^2}{x}=\frac{y}{2x}=f(x,y)$.

So the differential equation of the orthogonal family is $y'=-\frac{1}{f(x,y)}=-\frac{2x}{y}$

Solve this equation:

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

$ydy+2xdx=0$

$d\bigl(\frac{y^2}{2}+x^2\bigr)=0$

$\frac{y^2}{2}+x^2=C$

We obtain the family of ellipses $\frac{y^2}{2}+x^2=C=D^2$ or $\frac{x^2}{D^2}+\frac{y^2}{2D^2}=1$ $\frac{x^2}{D^2}+\frac{y^2}{2D^2}=1$

Answer: $\frac{x^2}{D^2}+\frac{y^2}{2D^2}=1$