Question Number 128430 by sachin1221 last updated on 07/Jan/21
$$\boldsymbol{{if}}\:\boldsymbol{{we}}\:\boldsymbol{{have}}\:\boldsymbol{{DE}} \\ $$$$\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{x}}\left({t}\right)}{\boldsymbol{{dt}}^{\mathrm{2}} }\:=\:−\boldsymbol{{sinx}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{find}}\:\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right) \\ $$$$\mathrm{1}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{acosht}}+\boldsymbol{{bsinht}} \\ $$$$\mathrm{2}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{a}}+\boldsymbol{{bt}} \\ $$$$\mathrm{3}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{ae}}^{\boldsymbol{{t}}} \:+\:\boldsymbol{{be}}^{\mathrm{2}\boldsymbol{{t}}} \\ $$$$\mathrm{4}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{acost}}\:+\:\boldsymbol{{bsint}} \\ $$
Commented by mr W last updated on 07/Jan/21
$${i}\:{think}\:{all}\:{options}\:{are}\:{wrong}. \\ $$
Commented by sachin1221 last updated on 07/Jan/21
$$\boldsymbol{{then}}\:\boldsymbol{{what}}\:\boldsymbol{{is}}\:\boldsymbol{{correct}}\:\boldsymbol{{answer}} \\ $$$$ \\ $$
Answered by mr W last updated on 07/Jan/21
$${let}\:{x}'={u} \\ $$$${x}''={u}\frac{{du}}{{dx}} \\ $$$${u}\frac{{du}}{{dx}}=−\mathrm{sin}\:{x} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{cos}\:{x}+{a} \\ $$$${u}=\frac{{dx}}{{dt}}=\pm\sqrt{\mathrm{2}\left(\mathrm{cos}\:{x}+{a}\right)} \\ $$$$\frac{{dx}}{\:\sqrt{\mathrm{cos}\:{x}+{a}}}=\pm\sqrt{\mathrm{2}}{dt} \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{cos}\:{x}+{a}}}=\pm\sqrt{\mathrm{2}}\int{dt} \\ $$$$\frac{\mathrm{2}{F}\left(\frac{{x}}{\mathrm{2}}\mid\frac{\mathrm{2}}{{a}+\mathrm{1}}\right)}{\:\sqrt{{a}+\mathrm{1}}}=\sqrt{\mathrm{2}}\left({b}\pm{t}\right) \\ $$$${F}\left(\frac{{x}}{\mathrm{2}}\mid\frac{\mathrm{2}}{{a}+\mathrm{1}}\right)=\left({b}\pm{t}\right)\sqrt{\frac{{a}+\mathrm{1}}{\mathrm{2}}} \\ $$$${or} \\ $$$${F}\left(\frac{{x}}{\mathrm{2}}\mid\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)={a}\left({b}\pm{t}\right) \\ $$$${F}={incomplete}\:{elliptic}\:{integral}\:{of} \\ $$$${the}\:{first}\:{kind} \\ $$
Commented by sachin1221 last updated on 10/Jan/21
$${i}\:{didn}'{t}\:{get}\:{last}\:\mathrm{3}\:{steps} \\ $$
Commented by mr W last updated on 10/Jan/21
