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Question Number 157267 by cortano last updated on 21/Oct/21
$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}={x}\: \\ $$
Commented by MathsFan last updated on 22/Oct/21
$${any}\:{caption}???? \\ $$
Commented by john_santu last updated on 22/Oct/21
$${y}={c}_{\mathrm{1}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)+{c}_{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{x}\left(\mathrm{1}+\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$
Commented by tabata last updated on 23/Oct/21
$${how}\:???? \\ $$