1/4 y^{''}+4y=tan(4t)-1/2 e^{2t}

解

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{2} e^{2 t}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 cos ( 4 t ) ln ( tan ( 4 t ) + sec ( 4 t ) ) - 2 e ^{2 t} cos ^{2} ( 4 t ) - 2 e ^{2 t} sin ^{2} ( 4 t )}{20}

解答ステップ

\frac{1}{4} y^{''} (t) + 4 y = tan (4 t) - \frac{1}{2} e^{2 t}

線形 ODE を解く:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 cos ( 4 t ) ln ( tan ( 4 t ) + sec ( 4 t ) ) - 2 e ^{2 t} cos ^{2} ( 4 t ) - 2 e ^{2 t} sin ^{2} ( 4 t )}{20}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 cos ( 4 t ) ln ( tan ( 4 t ) + sec ( 4 t ) ) - 2 e ^{2 t} cos ^{2} ( 4 t ) - 2 e ^{2 t} sin ^{2} ( 4 t )}{20}