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Question Number 190259 by jlewis last updated on 30/Mar/23
solve the differential   equation.  (d^2 /dt^2 ) x + ω^2 x(t) =0  ;x(0)=0;x^2 (0)=υ_o
$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\: \\ $$$$\mathrm{equation}. \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dt}^{\mathrm{2}} }\:\mathrm{x}\:+\:\omega^{\mathrm{2}} \mathrm{x}\left(\mathrm{t}\right)\:=\mathrm{0} \\ $$$$;\mathrm{x}\left(\mathrm{0}\right)=\mathrm{0};\mathrm{x}^{\mathrm{2}} \left(\mathrm{0}\right)=\upsilon_{\mathrm{o}} \\ $$
Commented by mr W last updated on 30/Mar/23
x(t)=(v_0 /ω) sin ωt
$${x}\left({t}\right)=\frac{{v}_{\mathrm{0}} }{\omega}\:\mathrm{sin}\:\omega{t} \\ $$
Commented by mehdee42 last updated on 30/Mar/23
x^2 (0)=0≠v_0
$${x}^{\mathrm{2}} \left(\mathrm{0}\right)=\mathrm{0}\neq{v}_{\mathrm{0}} \\ $$
Commented by mr W last updated on 31/Mar/23
x(0)=0 and x^2 (0)=v_0  is non−sense.  it is meant x′(0)=v_0 , see question  below.
$${x}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:{x}^{\mathrm{2}} \left(\mathrm{0}\right)={v}_{\mathrm{0}} \:{is}\:{non}−{sense}. \\ $$$${it}\:{is}\:{meant}\:{x}'\left(\mathrm{0}\right)={v}_{\mathrm{0}} ,\:{see}\:{question} \\ $$$${below}. \\ $$
Commented by mehdee42 last updated on 31/Mar/23
exactly.you are right.it was strang for mr too.  the question is typed incorrectly.thank you very much
$${exactly}.{you}\:{are}\:{right}.{it}\:{was}\:{strang}\:{for}\:{mr}\:{too}. \\ $$$${the}\:{question}\:{is}\:{typed}\:{incorrectly}.{thank}\:{you}\:{very}\:{much} \\ $$

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