Differential Equations Question - None

Question

The orthogonal trajectories of the family of curves $a^{n-1}y = x^n$ are given by (a is the arbitrary constant)

Accepted Answer

Step 1: Find the differential equation of the given family.

Given: $a^{n-1}y = x^n$

Differentiating with respect to x:

$a^{n-1}\frac{dy}{dx} = nx^{n-1}$

Therefore: $\frac{dy}{dx} = \frac{nx^{n-1}}{a^{n-1}}$

Step 2: Eliminate the arbitrary constant a.

From the original equation: $a^{n-1} = \frac{x^n}{y}$

Substituting into the derivative:

$\frac{dy}{dx} = \frac{nx^{n-1}}{\frac{x^n}{y}} = \frac{nxy}{x^n} = \frac{ny}{x}$

Step 3: Find the differential equation of orthogonal trajectories.

For orthogonal trajectories, replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$:

$-\frac{dx}{dy} = \frac{ny}{x}$

Therefore: $\frac{dx}{dy} = -\frac{ny}{x}$

Rearranging: $x\,dx = -ny\,dy$

Step 4: Integrate both sides.

$\int x\,dx = -n\int y\,dy$

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

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

Multiplying by 2:

$x^2 + ny^2 = ext{constant}$

∴ Answer: a

Question: