Question Number 169333 by Mastermind last updated on 28/Apr/22
$${Show}\:{that}\:{if}\:{y}={C}_{\mathrm{1}} {sinx}\:+\:{C}_{\mathrm{2}} {x}\:{then} \\ $$$$\left(\mathrm{1}+{xcotx}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−{x}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$ \\ $$$${Mastermind} \\ $$
Commented by som(math1967) last updated on 29/Apr/22
$${I}\:{think}\:\left(\mathrm{1}+{xcotx}\right)\:{should}\:{be} \\ $$$$\:\left(\mathrm{1}−\boldsymbol{{xcotx}}\right) \\ $$
Commented by Mastermind last updated on 29/Apr/22
$${if}\:{it}\:{is}\:−\:,\:{continue}\:{solving}\:{it} \\ $$
Commented by som(math1967) last updated on 29/Apr/22
$$\boldsymbol{{y}}={C}_{\mathrm{1}} {sinx}+{C}_{\mathrm{2}} {x} \\ $$$$\:\frac{{dy}}{{dx}}={C}_{\mathrm{1}} {cosx}\:+{C}_{\mathrm{2}} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=−{C}_{\mathrm{1}} {sinx} \\ $$$${now} \\ $$$$\left(\mathrm{1}−{xcotx}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−{x}\frac{{dy}}{{dx}}\:+{y} \\ $$$$\left(\mathrm{1}−{x}\frac{{cosx}}{{sinx}}\right)\left(−{C}_{\mathrm{1}} {sinx}\right)−{x}\left({C}_{\mathrm{1}} {cosx}+{C}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:+{C}_{\mathrm{1}} {sinx}\:+{C}_{\mathrm{2}} {x} \\ $$$$=−{C}_{\mathrm{1}} {sinx}+{C}_{\mathrm{1}} {xcosx}−{C}_{\mathrm{1}} {xcosx}−{xC}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+{C}_{\mathrm{1}} {sinx}\:+{C}_{\mathrm{2}} {x} \\ $$$$=\mathrm{0} \\ $$$$\:\: \\ $$