Find the values of a and b such that ys(x) = ax + b is a solution to the ordinary differential equation (ODE) Find the values of $a$ and $b$ such that $y_s(x) = ax + b$ is a solution to the ordinary differential equation (ODE)
$$2y''(x)=\frac{(y'(x))^{2}}{y(x)}$$
 A: Try $a=0, b=3$ for example. $b$ could be any number really. Just plug it into the equation.
If $y(x)=ax+b$ then  $y'(x)=a$ and $y''(x)=0$. Plug in all the numbers to get:
$$ 0 = \frac{a^2}{ax+b}$$
This is solved if we set $a=0$ it doesn't matter what $b$ is. (Assuming we are only allowed finite numbers).
A: If
$y(x) = ax+b$,
then
$y'(x) = a$
and
$y''(x) = 0$.
Therefore
we must have
$0
=\frac{(y'(x))^{2}}{y(x)}
=\frac{a^2}{ax+b}
$.
This can only hold if
$a=0$,
so that
$y(x) = b$.
If $b=0$,
then
$\frac{(y'(x))^{2}}{y(x)}$
is not defined,
so we must have
$b \ne 0$.
If
$b \ne 0$,
then
$y(x) = b$
satisfies
$y''(x)
=\frac{(y'(x))^{2}}{y(x)}
$
since both sides are zero.
Therefore,
that is the general solution.
A: The answer of the question was already given. So, this isn't a direct answer, but additional information.
$$2y''(x)=\frac{(y'(x))^{2}}{y(x)}$$
Obvious trivial solution : $y'=0 \quad\to\quad y(x)=$constant.
General solution with $y'\neq 0$ :
$$2\frac{y''}{y'}=\frac{y'}{y} \quad\to\quad (y')^2=cy\quad\to\quad \frac{y'}{\sqrt{\pm y}}=c$$
$$\sqrt{\pm y}=\frac{\sqrt{c}}{2}x+c_2=c_1x+c_2$$
The general solution is :
$$y(x)=\pm(c_1x+c_2)^2$$
This general solution includes the trivial solution $y(x)$=constant in case of $c_1=0$. 
Looking for solution on the form $y(x)=ax+b$ implies that the quadratic function $\pm(c_1x+c_2)^2$ reduces to the linear function $ax+b$. This is possible only if $c_1=0 \quad\to\quad$
$\begin{cases} a=0 \\ b=\pm (c_2)^2 \end{cases}$ . 
This is consistent with the previous answers, except that the general solution of the ODE  $\quad y(x)=\pm(c_1x+c_2)^2\quad$ was not mentioned.
