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

解答

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 12 cos ( 2 t ) ln ( tan ( 2 t ) + sec ( 2 t ) ) - e ^{2 t} cos ^{2} ( 2 t ) - e ^{2 t} sin ^{2} ( 2 t )}{8}

求解步骤

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

求解线性方程:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 12 cos ( 2 t ) ln ( tan ( 2 t ) + sec ( 2 t ) ) - e ^{2 t} cos ^{2} ( 2 t ) - e ^{2 t} sin ^{2} ( 2 t )}{8}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 12 cos ( 2 t ) ln ( tan ( 2 t ) + sec ( 2 t ) ) - e ^{2 t} cos ^{2} ( 2 t ) - e ^{2 t} sin ^{2} ( 2 t )}{8}

y^{^{''}} + 4 y^{^{'}} + y = - 9

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 6

y^{^{''}} + y = 10

4 y^{^{''}} + y = e^{- 2 t} cos (\frac{t}{2}) + 6 t sin (\frac{t}{2})

l in e a r y^{^{''''}} - 2 y^{^{'''}} + y^{^{''}} = t^{3}