Finding a differential equation

Question

How do I find a differential equation for this equation:

$ax+(y-b)^2=0$

I've tried deriving for $x$ and $y$ but it didn't work out very well.

Answer 1

The trick is to get the constants by themselves, so they go away after differentiation.

Solve for $a$ and differentiate

$$ \frac{(y-b)^2}{x} = a $$ $$ \frac{2(y-b)y'x - (y-b)^2}{x^2} = 0$$ $$ 2xy' - (y-b) = 0 $$

Solve for $b$ and differentiate again

$$ 2xy' - y = b $$ $$ 2xy'' + 2y' - y' = 0 $$ $$ \boxed{2xy'' + y' = 0} $$

You can check by solving the equation $$ \frac{y''}{y'} = -\frac{1}{2x} $$ $$ \ln{y'} = C -\frac{1}{2}\ln x $$ $$ y' = c_1x^{-1/2} $$ $$ y = c_2 + 2c_1x^{1/2} $$ $$ 4{c_1}^2x = (y-c_2)^2 $$

or $4{c_1}^2 = a$ and $c_2 = b$

Answer 2

So, we start from here

$ \begin{eqnarray} ax+(y−b)^2=0,\hspace{3cm}(1) \end{eqnarray} $

where $a$ and $b$ are constants and $y(x)$, then:

Taking a first implicit derivative

$ \begin{eqnarray} a+2y'(y-b)=0\hspace{3cm}(2) \end{eqnarray} $

One more derivative:

$ \begin{eqnarray} 2y''(y-b)+2(y')^2=0\hspace{3cm}(3) \end{eqnarray} $

From (2) we got:

$ \begin{eqnarray} a=-2y'(y-b)\hspace{3cm}(4) \end{eqnarray} $

Replacing (4) in (1) we got:

$ \begin{eqnarray} -2y'(y-b)x+(y−b)^2=0\hspace{3cm} \end{eqnarray} $

If $y\neq b$ then

$ \begin{eqnarray} -2y'x+(y−b)=0\hspace{3cm}(5) \end{eqnarray} $

and from here $(y-b)=2xy'$, which leaves (3) as:

$ \begin{eqnarray} 4y''xy'+2(y')^2=0\hspace{3cm} \end{eqnarray} $

or

$ \begin{eqnarray} 2y'(2y''x+y')=0\hspace{3cm}(6) \end{eqnarray} $

This is a general solution for the problem (no matter if $y=b$). When $y=b$, you have that $y'=0$ and when $y\neq b$ you have $2y''x+y'=0$.