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Beliebt Rechnen >

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

Lösung

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

Lösung

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

Graph

Beliebte Beispiele

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

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

y^{^{''}} - 9 y = 24

(d^2y)/(dt^2)-6(dy)/(dt)+9y=t

\frac{d ^{2} y}{d t ^{2}} - 6 \frac{d y}{d t} + 9 y = t

y^{''}-y^'-2y=3cos(2x)

y^{^{''}} - y^{^{'}} - 2 y = 3 cos (2 x)

6y^{'''}+4y^'=e^xcos(2x)

6 y^{^{'''}} + 4 y^{^{'}} = e^{x} cos (2 x)