How to solve $ \frac{d^2y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$

Question

I'm trying to solve:

$ \frac{d^2y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$

I've been following a tutorial and they say the answer is:

$y = \frac{1}{2}Cx^2 + D$

They've used:

$\frac{dy}{dx} = Cx$

How did they get that result??

Answer 1

I think that the equation read as follows:

$\frac{d^2 y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$. Put $z: =y'$ then we have

$z'=\frac{1}{x}z$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = \frac{1}{2}Cx^2 + D$.

Answer 2

Let us consider your differential equation: $$y''(x)-\frac{1}{x}\cdot y'(x)=0.$$ Add $\frac{1}{x}\cdot y'(x)$ on both sides: $$y''(x)=\frac{1}{x}\cdot y'(x).$$ Divide by $y'(x)$ on both sides: $$\frac{y''(x)}{y'(x)}=\frac{1}{x}.$$ Integrate with respect to $x$ on both sides: $$\int \frac{y''(x)\ dx}{y'(x)}=\int \frac{dx}{x}.$$ It follows that $$\ln\left(y'(x)\right)=C+\ln(x).$$ An expression for $y'(x)$ is given by $$y'(x)=e^{C+\ln(x)}=e^{C}\cdot e^{\ln(x)}=e^{C}\cdot x.$$ Redefine $e^{C}$ as $C$, since it is an arbitrary constant: $$y'(x)=C\cdot x.$$ Integrate with respect to $x$ on both sides: $$\int y'(x)\ dx=\int C\cdot x\ dx.$$ An expression for $y(x)$ is given by $$y(x)=\frac{1}{2}\cdot C\cdot x^2 + D.$$

Answer 3

The equation writes

$$\frac{y''(x)}{y'(x)}=\frac1x$$

or after integration

$$\frac{y'(x)}{y'_0}=\frac x{x_0}.$$

Then after a second integration,

$$\frac{y(x)-y_0}{y'_0}=\frac{x^2-x_0^2}{2x_0}$$

or

$$y(x)=\frac{y'_0}{2x_0}x^2+y_0-\frac{y'_0x_0^2}{2x_0}.$$