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

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.

• There is a standard technique of reducing order for $f(y,y',y'')=0$, that is putting $y'=p(y)$, then $y''=p\cdot p'_y$ Commented Mar 5, 2022 at 3:48

$$y y''=3(y')^2$$ 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.

• Of course it was the constant function..... shaking fist in the air. Why do I always fail to consider the constant fucntion. Thank you... Commented Mar 4, 2022 at 16:11
• You're welcome @Mando Commented Mar 4, 2022 at 16:15

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.

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, $$y=\frac{\gamma}{\sqrt{x_0-x}}\\ \gamma =\pm \sqrt{\frac{2y_0^3}{y^{'}_0}}$$

• Somehow the last step is missing constants, when compared with $$y^{-2}-y_0^{-2}=-2\frac{y_0'}{y_0^3}(x-x_0).$$ Commented Mar 4, 2022 at 23:03