if $y_1=\frac{\sin x}{\sqrt x}$ is solution of $x^2y''+xy'+(x^2-1/4)y=3 \sqrt x^3 \sin x$, find the other lineraly independent solutions

Question

If $y_1=\displaystyle\frac{\sin x}{\sqrt x}$ is a solution of the differential equation $$x^2y'' + xy' + \left(x^2-\frac{1}{4}\right)y=3\sqrt x^3 \sin x,$$ find the other linearly independent solution.

Answer 1

If $y_1(x)$ is obne solution of $y''+p(x)y'+qy=0$, then its other solution is $$y_2(x)=y_1(x)\int \frac{e^{\int -p(x) dx}}{y_1^2} dx$$ Please chcek that $y_1=\frac{\sin x}{\sqrt{x}}$ is also a solution of the HOMOGENEOUS ODE: $$x^2y''+xy'+(x^2-1/4)y=0\implies y''+\frac{1}{x}y'+(1-\frac{1}{4x^2})y=0,$$ So its other solution is given as $$y_2(x)=y_1(x)\int x \frac{e^{-\ln x}}{\sin^2 x} dx= y_1(x) \int \csc^2 x dx=-\frac{\sin x}{\sqrt{x}} \cot x= -\frac{\cos x}{\sqrt{x}}.$$

Further the solution of the inhomogeneous eq. of OP can be solved by the method of variation of parameters by seeking $$Y(x)=C_1(x) y_1(x)+ C_2(x) y_2(x)+D_1 y_1(x)+D_2 y_2(x)$$

See:https://en.wikipedia.org/wiki/Variation_of_parameters

Answer 2

Set, inspired by the given solution and the form of the right side, $u(x)=\sqrt{x}y(x)$. Then its derivatives are \begin{align} u'(x)&=\sqrt{x}y'(x)+\frac1{2\sqrt x}y(x)\\ u''(x)&=\sqrt{x}y''(x)+\frac1{\sqrt x}y'(x)-\frac1{4\sqrt x^3}y(x) \end{align} so that inserted into the given equation $$ \sqrt x^3u''(x)=x^2y''(x)+xy'(x)-\frac14y(x)=\sqrt x^3(3\sin x - u(x)) $$ Now cancelling again the common factor one is left with $$ u''(x)+u(x)=3\sin(x) $$ which is a standard forced harmonic oscillator (with resonance).