-y^{''}+y=e^{a*x},y(0)=0,y(1)=0

解

- y^{^{''}} + y = e^{a \cdot x}, y (0) = 0, y (1) = 0

y = e^{x} (- \frac{- e ^{a + 1} + e ^{2}}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} - \frac{1}{- a ^{2} + 1}) + \frac{e ^{- x} ( - e ^{a + 1} + e ^{2} )}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} + \frac{e ^{a x}}{- a ^{2} + 1}

解答ステップ

- y^{''} (x) + y = e^{a \cdot x}

線形 ODE を解く:

y = e^{x} (- \frac{- e ^{a + 1} + e ^{2}}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} - \frac{1}{- a ^{2} + 1}) + \frac{e ^{- x} ( - e ^{a + 1} + e ^{2} )}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} + \frac{e ^{a x}}{- a ^{2} + 1}

y = e^{x} (- \frac{- e ^{a + 1} + e ^{2}}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} - \frac{1}{- a ^{2} + 1}) + \frac{e ^{- x} ( - e ^{a + 1} + e ^{2} )}{( - e ^{2} + 1 ) ( - a ^{2} + 1 )} + \frac{e ^{a x}}{- a ^{2} + 1}