y^{''}-25y=0,y(2)=-2,y^'(-2)=2

解

y^{^{''}} - 25 y = 0, y (2) = - 2, y^{^{'}} (- 2) = 2

y = \frac{e ^{5 t} ( 2 e ^{10} + 5 e ^{20} \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} )}{5} + \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} e^{- 5 t}

解答ステップ

y^{''} (t) - 25 y = 0

線形 ODE を解く:

y = \frac{e ^{5 t} ( 2 e ^{10} + 5 e ^{20} \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} )}{5} + \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} e^{- 5 t}

y = \frac{e ^{5 t} ( 2 e ^{10} + 5 e ^{20} \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} )}{5} + \frac{- 10 e ^{10} - 2 e ^{30}}{5 ( e ^{40} + 1 )} e^{- 5 t}