Solve the ODE $y y''=3(y')^2$ using Reduction of Order

Question

For the reduction of order method, we are supposed to guess a solution, $y_1$, and assume that the second solution (this will be second order as the title shows) is of the form $y_2=uy_1$. But I'm having difficulty finding a solution by inspection. I've tried the following, which seemed legit at first, only to find that the constant 3 is making the problem more difficult.

$$y=e^3x \quad y=e^{\sqrt{3}x} \quad y=ax^2 \quad y=\sqrt{x}$$

Previously the problems given, it was pretty straight forward but this one is really hurting as I can't find a solution by inspection. It could be that we can maybe use some "product rule" manipulation, since

$$y y''=3(y')^2 \quad \rightarrow \quad y y''-3(y')^2=0$$

looks like it could be of the form $(y y')'$, but again, that 3 is messing me up. Once I find it, I want to try and find the other solution using Reduction of Order. So for anyone answering, please don't solve the ODE, just how I can maybe posit a guess given the information provided in the problem.

Answer 1

$$y y''=3(y')^2$$ Note that $y=C$ is also a solution. For $y\ne C$: $$y y''-y'^2=2(y')^2$$ $$\dfrac {y y''-y'^2}{(y')^2}=2$$ $$\left (\dfrac {y}{y'} \right)'=-2$$ Inetgrate to reduce the order of the DE.

Answer 2

You can first change from $y'$ to $x'$, then get a reduction of order. The inverse function formula for the second derivative is $y'' = -\left(\frac{1}{x'}\right)^3x''$. Therefore, your equation becomes:

$$-y\,\left(\frac{1}{x'}\right)^3x'' = 3\left(\frac{1}{x'}\right)^2$$

This reduces to:

$$-y\,\left(\frac{1}{x'}\right)x'' = 3$$

Now you have a problem where you can do reduction of order for $x$. So, $u = x'$ and $u' = x''$.

$$ -y\,\frac{1}{u}\,u' = 3 $$

Alternatively, you can write $u'$ as $\frac{du}{dy}$. So, this becomes:

$$ -y\,\frac{1}{u}\,\frac{du}{dy} = 3 \\ \frac{du}{u} = -3\frac{dy}{y} \\ \ln(u) = -3\,\ln(y) + C \\ u = \frac{C}{y^3} \\ \frac{dx}{dy} = \frac{C}{y^3} \\ dx = C y^{-3}\,dy \\ x = Cy^{-2} + D $$

Note that a few times in there, C got merged with another constant, but I didn't feel like spelling it out.

Answer 3

You can also write the d.e. $$ \frac {y^{"}}{y^{'}}=3\frac{y^{'}}{y}\\ \frac{dy^{'}}{y^{'}}=3\frac{dy}{y} $$ Integrate both sides, $$ \log(\frac{y^{'}}{y^{'}_0})=3\log(\frac{y}{y_0}) \\ y^{'}=\frac{y^{'}_0}{y_0^{3}}y^3\\ $$ Integrate again, $$ y=\frac{\gamma}{\sqrt{x_0-x}}\\ \gamma =\pm \sqrt{\frac{2y_0^3}{y^{'}_0}} $$