Find the differential equation corresponding to family of curves $y=c(x-c)^2$, where $c$ is an arbitrary constant
My attempt:
$$y=c(x-c)^2 \implies \sqrt y = \sqrt c (x-c)$$
Differentiating, $$\frac{y'}{2\sqrt y} = \sqrt c \implies c = \frac{(y')^2}{4 y}$$
Substituting value of $c$ in above given equation i get
then i am getting differential equation as
$$64y^4 = y'^2(4xy-y'^2)^2 \tag{1}$$
But if i differentiate above given equation
$$y'=2c(x-c) \implies c = x-\frac{2y}{y'}$$
With this i get
$$y'^3=4y(xy'-2y)\tag{2}$$
Now which one is correct – (1) or (2)?
Best Answer
Note that when we are given a differential equation containing $n$ arbitrary constants, by differentiating it successively $n$ times and eliminating the constants gives us the required differential equation of order $n$.
Here, $n=1$. Simplifying $(1)$ by taking the square root, we get: $$8y^2 = y’(4xy-y’^2) \implies y’^3=4xyy’-8y^2 = 4y(xy’-2y)$$ that is $(2)$.
Note that, simplifying $(1)$ gave us the simplest form of the differential equation. But, it is always preferred to differentiate the equation and eliminate constants.