# Second-Order Nonlinear Ordinary Differential Equation - with a scalar multiple

I studied differential equations back in the day but we never covered second-order nonlinear equations (that I can recall). I have the following equation:

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

Solving a second-order nonlinear ordinary differential equation

but the "2" out front is throwing me off.

I am not a student and this is not homework.

I am fairly certain that the first-order solution to this problem will be a Bernoulli type nonlinear equation and that much I can solve. It's just this 2 out front - and getting from a second-order form to a first-order form - that are throwing me.

Any help in solving for $y'$ would be greatly appreciated!

P. S. This is my first post. My apologies if there are any rules that I have failed to follow.

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Edit: The form of the first-order solution should have the form

$$y'(x)=P(x)y(x)+Q(x)y(x)^{2}$$

which would then be solved by the Bernoulli method. But how do we get from the second-order form to this one? I've been trying it on my own and nothing is working.

Thanks. Sorry for not being more specific.

• Start as in previous question so after first integration you have $y'=A\frac{y^2}{x^2}$ – user121049 Oct 17 '17 at 12:54

$$y''=2\biggl(\frac{y'^{2}}{y}-\frac{y'}{x}\biggr)$$ Factor out $y'$. Notice that $y'=0$ is one of the solutions. Anothere we can find from: $$\frac{y''}{y'}=2\frac{y'}{y}-2\frac{1}{x}$$ It can be rewritten as follows (use properties of logarithm): $$(\ln y')'=2(\ln y)'-2(\ln x)'\\ (\ln y')'=\left(\ln \left(\frac{y}{x}\right)^2\right)'$$ Integrate: $$\ln y'=\ln\left(\frac{y}{x}\right)^2+C$$ Exponentiate (and then you receive the equation specified in comments): $$y'=e^C \left(\frac{y}{x}\right)^2=A \left(\frac{y}{x}\right)^2$$
• $P(x)=0$, $Q(x)=A/x^2$ – user121049 Oct 17 '17 at 15:17
For this equation, define $y=\frac x z$ making the equation to be $$\frac {x z''}{z^2}=0\implies z''=0\implies z'=c_1\implies z=c_1x+c_2\implies y=\frac x{c_1x+c_2}$$