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Question: Form the differential equation of family of lines situated at a constant distance P from the origin...

Form the differential equation of family of lines situated at a constant distance P from the origin

Answer

P^2 (1 + (y')^2) = (y - xy')^2

Explanation

Solution

To form the differential equation for a family of lines, we first write down the general equation of such a family involving an arbitrary constant. Then, we differentiate this equation with respect to xx and eliminate the arbitrary constant using the original equation and its derivative.

  1. Equation of the family of lines:
    The equation of a straight line at a constant perpendicular distance PP from the origin is given by the normal form: xcos⁡α+ysin⁡α=P⋯(1)x \cos \alpha + y \sin \alpha = P \quad \cdots (1) Here, PP is a given constant, and α\alpha is the arbitrary constant (parameter) that varies for different lines in the family. Since there is one arbitrary constant (α\alpha), we will need to differentiate the equation once to form the differential equation.

  2. Differentiate with respect to x:
    Differentiating equation (1) with respect to xx: ddx(xcos⁡α+ysin⁡α)=ddx(P)\frac{d}{dx}(x \cos \alpha + y \sin \alpha) = \frac{d}{dx}(P) cos⁡α⋅ddx(x)+sin⁡α⋅ddx(y)=0\cos \alpha \cdot \frac{d}{dx}(x) + \sin \alpha \cdot \frac{d}{dx}(y) = 0 cos⁡α⋅1+sin⁡α⋅dydx=0\cos \alpha \cdot 1 + \sin \alpha \cdot \frac{dy}{dx} = 0 cos⁡α+y′sin⁡α=0⋯(2)\cos \alpha + y' \sin \alpha = 0 \quad \cdots (2) (where y′=dydxy' = \frac{dy}{dx})

  3. Eliminate the arbitrary constant α\alpha:
    From equation (2), we can express cos⁡α\cos \alpha in terms of sin⁡α\sin \alpha and y′y': cos⁡α=−y′sin⁡α\cos \alpha = -y' \sin \alpha

    Substitute this expression for cos⁡α\cos \alpha into equation (1): x(−y′sin⁡α)+ysin⁡α=Px(-y' \sin \alpha) + y \sin \alpha = P −xy′sin⁡α+ysin⁡α=P-xy' \sin \alpha + y \sin \alpha = P Factor out sin⁡α\sin \alpha: sin⁡α(y−xy′)=P\sin \alpha (y - xy') = P sin⁡α=Py−xy′\sin \alpha = \frac{P}{y - xy'}

    Now, substitute this expression for sin⁡α\sin \alpha back into the equation for cos⁡α\cos \alpha: cos⁡α=−y′(Py−xy′)=−Py′y−xy′\cos \alpha = -y' \left( \frac{P}{y - xy'} \right) = \frac{-Py'}{y - xy'}

    Finally, use the trigonometric identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1: (Py−xy′)2+(−Py′y−xy′)2=1\left( \frac{P}{y - xy'} \right)^2 + \left( \frac{-Py'}{y - xy'} \right)^2 = 1 P2(y−xy′)2+P2(y′)2(y−xy′)2=1\frac{P^2}{(y - xy')^2} + \frac{P^2 (y')^2}{(y - xy')^2} = 1 Combine the terms on the left side: P2(1+(y′)2)(y−xy′)2=1\frac{P^2 (1 + (y')^2)}{(y - xy')^2} = 1 Rearrange to get the differential equation: P2(1+(y′)2)=(y−xy′)2P^2 (1 + (y')^2) = (y - xy')^2

This is the required differential equation for the family of lines situated at a constant distance PP from the origin.