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

Lösung

- 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}

Schritte zur Lösung

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

Solve linear ODE:

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}

y^{^{'''}} - y^{^{''}} = x + e^{x}

y^{^{''}} + 4 y^{^{'}} + 32 y = 4 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{2} + 4}

y^{^{''}} + 4 y = 3 sin (t), y (0) = 1, y^{^{'}} (0) = - 1