Question

Помогите найти решение задачи Коши

Answer 1

$y''+18sin(y)cos^3(y)=0$

замена: $y'=p(y)$
$y''=pp'$

$pp'+18sin(y)cos^3(y)=0$

$pdp=-18sin(y)cos^3(y)dy$

$frac{p^2}{2}=-18intlimits {sin(y)cos^3(y)} , dx$

$frac{p^2}{2}=18intlimits {cos^3(y)} , d(cos(y))$

$frac{p^2}{2}=18*frac{cos^4(y)}{4}+C_0$

$p^2=9cos^4(y)+C$

$(y')^2=9cos^4(y)+C$

используем $y'(0)=3$

$9*1^4+C=3$

$(y')^2=9cos^4(y)-6$

$y'=pm sqrt{9cos^4(y)-6}$

$frac{dy}{sqrt{9cos^4(y)-6}}=pm dx$

$intlimits {frac{dy}{sqrt{9cos^4(y)-6}}} , dx=pm x+C$