Question

Find the orthogonal trajectories of the following families :

(i) $x^2 - \frac{1}{3}y^2 = a^2 \longrightarrow 2x - \frac{2}{3}y\frac{dy}{dx} = 0$

(ii) $y = \tan x + c \quad DE \quad of : x - \frac{y}{3}\frac{dy}{dx} = 0$

(iii) $\cos y = ae^{-x} \quad \text{family of curves}$

(iv) $y^2 = 4(x-a)$

$\downarrow \quad \text{replace} \frac{dy}{dx} \text{ by } -\frac{1}{\frac{dy}{dx}}$

Answer 1

The question asks to find the orthogonal trajectories for four families of curves. The solutions are: (i) $3x^2 + y^2 = C'$ (ii) $3x^2 + y^2 = C'$ (iii) $\sin y = A e^{-x}$ (iv) $y = A e^{-x/2}$

Answer 2

Here's a breakdown of how to find the orthogonal trajectories for each family:

Part (i) Family: $x^2 - \frac{1}{3}y^2 = a^2$

1. The given DE is $2x - \frac{2}{3}y\frac{dy}{dx} = 0$, which simplifies to $\frac{dy}{dx} = \frac{3x}{y}$.
2. For orthogonal trajectories, replace $\frac{dy}{dx}$ with $-\frac{1}{\frac{dy}{dx}}$, yielding $\frac{dy}{dx} = -\frac{y}{3x}$.
3. Solve by separation of variables: $y \, dy = -3x \, dx$.
4. Integrate: $\int y \, dy = \int -3x \, dx \implies \frac{y^2}{2} = -\frac{3x^2}{2} + C_1 \implies 3x^2 + y^2 = C'$.

Part (ii) Family defined by DE: $x - \frac{y}{3}\frac{dy}{dx} = 0$

1. The DE is $\frac{dy}{dx} = \frac{3x}{y}$.
2. For orthogonal trajectories, $\frac{dy}{dx} = -\frac{y}{3x}$.
3. Solve by separation of variables: $y \, dy = -3x \, dx$.
4. Integrate: $\int y \, dy = \int -3x \, dx \implies \frac{y^2}{2} = -\frac{3x^2}{2} + C_1 \implies 3x^2 + y^2 = C'$.

Part (iii) Family: $\cos y = ae^{-x}$

1. Differentiate: $-\sin y \frac{dy}{dx} = -ae^{-x}$. Substitute $ae^{-x} = \cos y$ to get $\sin y \frac{dy}{dx} = \cos y$, so $\frac{dy}{dx} = \cot y$.
2. For orthogonal trajectories, $\frac{dy}{dx} = -\frac{1}{\cot y} = -\tan y$.
3. Solve by separation of variables: $\frac{dy}{\tan y} = -dx \implies \cot y \, dy = -dx$.
4. Integrate: $\int \cot y \, dy = \int -dx \implies \ln|\sin y| = -x + C_1 \implies \sin y = A e^{-x}$.

Part (iv) Family: $y^2 = 4(x-a)$

1. Differentiate: $2y \frac{dy}{dx} = 4$, so $\frac{dy}{dx} = \frac{2}{y}$.
2. For orthogonal trajectories, $\frac{dy}{dx} = -\frac{1}{\frac{2}{y}} = -\frac{y}{2}$.
3. Solve by separation of variables: $\frac{dy}{y} = -\frac{1}{2} dx$.
4. Integrate: $\int \frac{dy}{y} = \int -\frac{1}{2} dx \implies \ln|y| = -\frac{1}{2}x + C_1 \implies y = A e^{-x/2}$.