Find a firstorder ODE in the form Fxyy 0 whose general solu

Find a first-order ODE in the form F(x,y,y) = 0, whose general solution is given by the family of parabolas y = cx2 + 2, where c is an arbitrary constant. (Hint: Calculate y and eliminate c in the two equations to find a relationship between y and y. That’s it!)

Solution

Given, y = cx²  + 2 ........................(1)

Differentiating equation (1) wrt \'x\' we get

y\' = c (2x) + 0 (Because derivative of x² is 2x and that of 2 is zero )

therefore y\' = (2x) c ......................(2)

from equation (2), c = y\' /(2x) .......................(3) (Because c is an arbitrary constant so we eliminate c )

substituting the value of c from equation (3) in equation (1) we get,

y = [y\'/ (2x)] x²  + 2

this implies y = [y\'/2]x + 2 (Because x²/x = x)

Multiplying by 2 on both sides we get,

2y = [ y\' ]x + 4

Therefore [ y\' ]x - 2y + 4 = 0 (transposing 2y to the right)

Therefore [ y\' ]x - 2y + 4 = 0 is the required ODE (ordinary differential equation)