y^{''}+y=cos^2(t),y(0)=0,y^'(0)=0

解

y^{^{''}} + y = cos^{2} (t), y (0) = 0, y^{^{'}} (0) = 0

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

解答ステップ

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

線形 ODE を解く:

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

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