The given equation is $$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
and the solution we are meant to verify is $$y_1(x)=x^{-1/2}\cos(x)$$ Taking the first and second derivative of this solution yields $$y^{'}_1(x)=-\frac{\sin(x)}{\sqrt{x}}-\frac{\cos(x)}{2x^{3/2}}$$ $$y^{''}_1(x)=-\frac{\cos(x)}{\sqrt{x}}+\frac{\sin(x)}{2x^{3/2}}+\frac{\cos(x)}{3x^{5/2}}+\frac{\sin(x)}{2x^{3/2}}$$ Now substituting in these derivative into their respective spots $$0=x^2\left(-\frac{\cos(x)}{\sqrt{x}}+\frac{\sin(x)}{x^{3/2}}+\frac{\cos(x)}{3x^{5/2}}\right)+x\left(-\frac{\sin(x)}{\sqrt{x}}-\frac{\cos(x)}{2x^{3/2}}\right)+(x^2-\frac{1}{4})\left(x^{-1/2}\cos(x)\right)$$ What I got as a final answer was $$-3x^{-1/2}\cos(x)\ne0$$
Am I correct or did I make a fault along the way?
This is good: $$y'_1(x)=-\frac{\sin(x)}{\sqrt{x}}-\frac{\cos(x)}{2x^{3/2}}$$ But this line is not correct: $$y''_1(x)=-\frac{\cos(x)}{\sqrt{x}}+\frac{\sin(x)}{2x^{3/2}}+\color{red}{\frac{\cos(x)}{3x^{5/2}}}+\frac{\sin(x)}{2x^{3/2}}$$ It should be: $$y''_1(x)=-\frac{\cos(x)}{\sqrt{x}}+\color{blue}{\dfrac 34\frac{\cos(x)}{x^{5/2}}}+\frac{\sin(x)}{x^{3/2}}$$