# How to solve $\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0$? [closed]

Let $y\in C^{2}(\mathbb R)$ (twice continuously differntiable function). We consider the ODE as follows: $$\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0$$ ($y$ is function of $x$)

My naive question is: (1) How to solve the above ODE? (2) How to make this question rigorous? (What can we say about solution, etc....)

## closed as off-topic by Travis, Bumblebee, TravisJ, Tim Raczkowski, Zain PatelJul 21 '15 at 16:56

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• Did you notice that there is no $y$ in equation? Just put $v=dy/dx$, and you get a separable first order equation. – orion Jul 21 '15 at 7:33

Write the ODE as $$xy''+y'=0$$ Where prime dentoes derivative wrt to $x$. Then notice the left hand side is simply the derivative of a product, so $$(xy')'=0$$ By implication therefore $$xy'=C$$ Where $C$ is a constant; thus $$dy=\frac{C dx}{x}$$

HINT: Try to make change of variables

\begin{aligned} u = \frac{dy}{dx} & \implies \frac{du}{dx}=\frac{d^{2}y}{dx^{2}} \implies \boxed{\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0 \iff \frac{d u}{dx} + \frac{1}{x} u =0} \end{aligned}

Note that the resulting differential equation $$\frac{d u}{dx} + \frac{1}{x} u = 0$$ is separable, so you can easily integrate it and find its solution $u = u(x)$, and consequently $y = y(x)$.

First note that $\color{blue}{y_1=1}$ is a solution, if you are looking for a second solution $y_2$ for an homogeneous ODE $$y''+p(x)y'+q(x)=0...(1)$$ you can use the formula $$y_2=y_1\int\frac{e^{-\int p(x)dx}}{y_1^2}dx$$ The general solution for $(1)$ can be written as $y(x)=c_1y_1+c_2y_2$ where $c_1$ and $c_2$ are constants.

• As far as theory goes - this is the best approach in my opinion. +1 – Autolatry Jul 21 '15 at 8:18

In general this is the Cauchy equation. The general method is to use the transformation $y=x^m$ for some integer $m$. Though for the present case if you simply take $p=dy/dx$ you'll get $$d(\ln p)/dx+1/x=0$$ which can be easily solved to obtain $p x=C\implies dy=C\frac{dx}{x}\implies y=C\ln x+D$.