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Find the values of $a$ and $b$ such that $y_s(x) = ax + b$ is a solution to the ordinary differential equation (ODE)

$$2y''(x)=\frac{(y'(x))^{2}}{y(x)}$$

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  • $\begingroup$ I got a=0, b=anything. $\endgroup$ – zooby Nov 18 '16 at 1:16
  • $\begingroup$ thats what my prof said but i dont understand how $\endgroup$ – ISuckAtMathPleaseHELPME Nov 18 '16 at 1:17
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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).

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  • $\begingroup$ Not only “if we set $a=0$”, but “only if $a=0$”. $\endgroup$ – Matthew Leingang Nov 18 '16 at 1:41
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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.

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  • $\begingroup$ Note : This is not the general solution of the PDE, but the particular solution according to the expected particular form of solution. Nevertheless, your answer is good. $\endgroup$ – JJacquelin Nov 18 '16 at 10:01
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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.

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