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Worked Examples · Example 6

Q.Form a differential equation representing the family of parabolas having vertex at origin and axis along positive direction of y-axis.

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A parabola with vertex at the origin and axis along +y+y is x2=4ayx^2=4ay; eliminating aa by differentiation gives xdydx=2yx\dfrac{dy}{dx}=2y.

A one-parameter family needs one differentiation to eliminate its single arbitrary constant, giving a first-order DE. Standard parabola about the yy-axis: x2=4ayx^2=4ay (aa = parameter).

Set up the family: vertex (0,0)(0,0), axis along the positive yy-axis ⇒\Rightarrow x2=4ayx^2=4ay, where a>0a>0 is arbitrary. ...(1)

  1. Differentiate (1) w.r.t. xx: 2x=4adydx  ⇒  4a=2x dy/dx 2x=4a\dfrac{dy}{dx}\;\Rightarrow\;4a=\dfrac{2x}{\,dy/dx\,}.
  2. From (1), 4a=x2y4a=\dfrac{x^2}{y}.
  3. Equate the two expressions for 4a4a: x2y=2x dy/dx \dfrac{x^2}{y}=\dfrac{2x}{\,dy/dx\,}.
  4. Cross-multiply: x2dydx=2xyx^2\dfrac{dy}{dx}=2xy.
  5. Divide by xx (for x≠0x\ne0): xdydx=2yx\dfrac{dy}{dx}=2y.
✓Final answer

  xdydx−2y=0  \;x\dfrac{dy}{dx}-2y=0\; (order 11, degree 11).

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