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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:

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

I tried to follow this

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.

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  • $\begingroup$ Start as in previous question so after first integration you have $y'=A\frac{y^2}{x^2}$ $\endgroup$
    – user121049
    Oct 17, 2017 at 12:54

2 Answers 2

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$$y''=2\biggl(\frac{y'^{2}}{y}-\frac{y'}{x}\biggr)$$ 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): $$ y'=e^C \left(\frac{y}{x}\right)^2=A \left(\frac{y}{x}\right)^2 $$

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  • $\begingroup$ Kostiantyn Lapchevskyi That isn't quite right. I followed what you did (I think) this isn't the form I think we should have. Sorry, I should have been more clear in the original post. I will edit it. $\endgroup$
    – user492494
    Oct 17, 2017 at 15:10
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    $\begingroup$ $P(x)=0$, $Q(x)=A/x^2$ $\endgroup$
    – user121049
    Oct 17, 2017 at 15:17
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For this equation, define $y=\frac x z$ making the equation to be $$\frac {x z''}{z^2}=0\implies z''=0\implies z'=c_1\implies z=c_1x+c_2\implies y=\frac x{c_1x+c_2}$$

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