y^{''}+y=sin^2(t)

解

y^{^{''}} + y = sin^{2} (t)

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) (cos (t) - \frac{1}{3} cos^{3} (t)) + \frac{sin ^{4} ( t )}{3}

解答ステップ

y^{''} (t) + y = sin^{2} (t)

線形 ODE を解く:

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) (cos (t) - \frac{1}{3} cos^{3} (t)) + \frac{sin ^{4} ( t )}{3}

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) (cos (t) - \frac{1}{3} cos^{3} (t)) + \frac{sin ^{4} ( t )}{3}