k4s3thg368

2022-03-24

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

Makenzie Hart

Step 1
To find the orthogonal family of curves, you need to first solve for c in terms of x and y.
${y}^{2}=c{e}^{x}-x-1$
$2y\sim {y}^{\prime }=c{e}^{x}-1$
From the given family of curves, $c=\frac{1+x+{y}^{2}}{{e}^{x}}$
So, ${y}^{\prime }=\frac{x+{y}^{2}}{2y}$
Slope for the orthogonal trajectories will be $\sim {y}^{\prime }=-\frac{2y}{x+{y}^{2}}$
Or, , which is of the form .
Step 2
To solve the above differential equation, note that it is not exact $\left({M}_{y}\ne {N}_{x}\right)$ so first find an integrating factor that makes it exact, which in this case is $\frac{1}{\sqrt{y}}$.
Multiplying by the integrating factor,

As ${M}_{y}={N}_{x}$, it is now exact and integrating, we get the solution
$2x\sqrt{y}+{\frac{25}{y}}^{\frac{5}{2}}=C$
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.

Do you have a similar question?