Question Number 144930 by alcohol last updated on 30/Jun/21
$${solve} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }={cosx} \\ $$
Answered by ArielVyny last updated on 30/Jun/21
$$−{cosx} \\ $$
Answered by mr W last updated on 01/Jul/21
$${u}=\frac{{dx}}{{dt}} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }=\frac{{du}}{{dt}}={u}\frac{{du}}{{dx}}=\mathrm{cos}\:{x} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{sin}\:{x}+{C} \\ $$$${u}=\frac{{dx}}{{dt}}=\pm\sqrt{\mathrm{2}\left(\mathrm{sin}\:{x}+{C}\right)} \\ $$$$\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}+{C}}}=\pm\sqrt{\mathrm{2}}{dt} \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}+{C}}}=\pm\sqrt{\mathrm{2}}\int{dt} \\ $$$$\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}+{C}}}=\pm\sqrt{\mathrm{2}}{t}+{C}_{\mathrm{1}} \\ $$