1/2 y^{''}+2y=4tan(2t)-1/3 e^t

解答

\frac{1}{2} y^{^{''}} + 2 y = 4 tan (2 t) - \frac{1}{3} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 30 cos ( 2 t ) ln ( sec ( 2 t ) + tan ( 2 t ) ) - 2 e ^{t} cos ^{2} ( 2 t ) - 2 e ^{t} sin ^{2} ( 2 t )}{15}

求解步骤

\frac{1}{2} y^{''} (t) + 2 y = 4 tan (2 t) - \frac{1}{3} e^{t}

求解线性方程:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 30 cos ( 2 t ) ln ( sec ( 2 t ) + tan ( 2 t ) ) - 2 e ^{t} cos ^{2} ( 2 t ) - 2 e ^{t} sin ^{2} ( 2 t )}{15}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 30 cos ( 2 t ) ln ( sec ( 2 t ) + tan ( 2 t ) ) - 2 e ^{t} cos ^{2} ( 2 t ) - 2 e ^{t} sin ^{2} ( 2 t )}{15}

38 y^{^{''}} - 7 y = \frac{5}{2}

3 y^{^{''}} + 12 y = - 4 sin (3 t)

y^{^{'''}} + 4 y^{^{'}} = 3 e^{x} (2 x + 5)

y^{^{''''}} - 4 y^{^{''}} = 5 x^{2} - e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 8} e^{2 t}