find a second solution y2(x) ofx^2y"-3xy'+5y=0; y1=x^2cos(lnx)

We can try reduction order and look for a solution [tex]y_2(x)=y_1(x)v(x)[/tex]. Then[tex]y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v[/tex]Substituting these into the ODE gives[tex]x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0[/tex][tex]x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0[/tex][tex]x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0[/tex]which leaves us with an ODE linear in [tex]w(x)=v'(x)[/tex]:[tex]x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0[/tex]This ODE is separable; divide both sides by the coefficient of [tex]w'(x)[/tex] and separate the variables to get[tex]w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0[/tex][tex]\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}[/tex][tex]\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx[/tex]Integrate both sides; on the right, substitute [tex]u=\ln x[/tex] so that [tex]\mathrm du=\dfrac{\mathrm dx}x[/tex].[tex]\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du[/tex]Now solve for [tex]w(u)[/tex],[tex]\ln|w|=-2\ln(\cos u)-u+C[/tex][tex]w=e^{-2\ln(\cos u)-u+C}[/tex][tex]w=Ce^{-u}\sec^2u[/tex]then for [tex]w(x)[/tex],[tex]w=Ce^{-\ln x}\sec^2(-\ln x)[/tex][tex]w=C\dfrac{\sec^2(\ln x)}x[/tex]Solve for [tex]v(x)[/tex] by integrating both sides.[tex]v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx[/tex]Substitute [tex]u=\ln x[/tex] again and solve for [tex]v(u)[/tex]:[tex]v=\displaystyle C_1\int\sec^2u\,\mathrm du[/tex][tex]v=C_1\tan u+C_2[/tex]then for [tex]v(x)[/tex],[tex]v=C_1\tan(\ln x)+C_2[/tex]So the second solution would be[tex]y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)[/tex][tex]y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)[/tex][tex]y_1(x)[/tex] already accounts for the second term of the solution above, so we end up with[tex]\boxed{y_2=x^2\sin(\ln x)}[/tex]as the second independent solution.