Given $y^2=ce^x-x-1$ Find the orthogonal family of curves and write 2 curves from both families that pass in (2,0).

Solution 1:

To find the orthogonal family of curves, you need to first solve for $c$ in terms of $x$ and $y$.

$ \displaystyle y^2=ce^x-x-1$

$2 y ~y' = ce^x - 1$

From the given family of curves, $c = \frac{1 + x + y^2}{e^x}$

So, $y' = \frac{x + y^2}{2y}$

Slope for the orthogonal trajectories will be $ ~y' = - \frac{2y}{x + y^2}$

Or, $ ~2y ~dx + (x+y^2) ~ dy = 0$, which is of the form $M ~ dx + N ~ dy = 0$.

To solve the above differential equation, note that it is not exact $(M_y \ne N_x)$ so first find an integrating factor that makes it exact, which in this case is $ \frac{1}{\sqrt y}$.

Multiplying by the integrating factor,

$2 \sqrt y ~ dx + \left(\frac{x}{\sqrt y} + y^{3/2}\right) dy = 0$

As $M_y = N_x$, it is now exact and integrating, we get the solution

$2 x \sqrt y + \frac 25 y^{5/2} = C$

To get two curves from both families that pass through the point $(2, 0)$, plug in $x = 2, y = 0$ in the equation of curves and find values of $c$ and $C$.