# Knowing the particular solution $y_1 = \frac{\sin(x)}{x}$ solve: $y'' + \frac{2}{x}y' + y = 0$

Knowing the particular solution $$y_1 = \frac{\sin(x)}{x}$$ solve: $$y'' + \frac{2}{x}y' + y = 0$$

I set the substitution $$y = vy_1 = v\frac{\sin(x)}{x}$$. I get the following:

$$y' = v' \frac{\sin(x)}{x} +v\frac{x\cos(x) - \sin(x)}{x^2}$$ $$y'' = v''\frac{\sin(x)}{x} + 2v'\frac{\cos(x)}{x}-2v'\frac{\sin(x)}{x^2}-2v\frac{\cos(x)}{x^2}-v\frac{\sin(x)}{x}-v\frac{2\sin(x)}{x^3}$$

Which results in:

$$v'' + v'(2\cot(x)) = \frac{4}{x^2}$$

Now, this linear differential is not very clean, and I am not sure if there is an error somewhere, or if this approach is invalid.

$$y'' + \frac{2}{x}y' + y = 0$$ The DE is simply: $$(xy)''+xy=0$$ $$\implies r^2+1=0 \implies r=\pm i$$ $$xy=c_1\cos x +c_2\sin x$$ $$y(x)=\dfrac 1 x (c_1\cos x +c_2\sin x)$$
You made a little mistake you should get: $$(xy)''+xy=0$$ $$(v\sin x)''+v \sin x=0$$ $$v'' \sin x +2v' \cos x=0$$ $$v''+2 v'\cot x=0$$