Second-order linear ordinary differential equation: $x^2y'' - 2xy = 0$

Question

$$x^2y'' - 2xy = 0$$ Professor has recently given this ODE as an assignment. It does look like a Bessel differential equation, however, I was not able to proceed with solution. What are the steps?

Answer 1

Our objective is to get our ODE in the following form (Modified Bessel Differential Equation): $$x^2y''(x)+xy'(x)-(x^2+n^2)y(x)=0 \tag{1}$$ This ODE has a well known general solution, namely: $$y(x)=c_1I_n(x)+c_2K_n(x) \tag{2}$$ Where $I_n(x)$ and $K_n(x)$ are modified Bessel functions of the first and second kind respectively.

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We start with the ODE you are given: $$x^2\cdot y''(x)-2x\cdot y(x)=0$$

Substituting $u=\sqrt{8x}$, or $x=\dfrac{u^2}{8}$ yields the following via the chain rule and product rule: $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{4}{u} \qquad \text{and} \qquad \frac{d^2y}{dx^2}=\frac{d^2y}{du^2}\cdot \frac{16}{u^2}-\frac{dy}{du}\cdot \frac{16}{u^3}$$ Thus, we obtain the following ODE (I'll leave the checking as an exercise): $$u^2 y''(u)-uy'(u)-u^2y(u)=0$$ Finally, substituting $y=uv$ gives the following via the product rule: $$\frac{dy}{du}=v+u\frac{dv}{du} \qquad \text{and} \qquad \frac{d^2y}{du^2}=2\frac{dv}{du}+u\frac{d^2v}{du^2}$$ Thus, we obtain the following ODE (Again, I'll leave the checking as an exercise): $$u^2v''(u)+uv'(u)-(u^2+1)v(u)=0 \tag{3}$$ We see that equation $(3)$ is in the same form as $(1)$ with $n=1$. Thus, substituting back for $x$ and $y$, we obtain the following solution: $$y(x)=c_1\sqrt{x}I_1(\sqrt{8x})+c_2\sqrt{x}K_1(\sqrt{8x})$$