y^{''}-4y=0,y^'(-1)=1,y(2)=2

Lösung

y^{^{''}} - 4 y = 0, y^{^{'}} (- 1) = 1, y (2) = 2

y = \frac{( e ^{2} ( e ^{12} + 1 ) + e ^{4} ( 4 e ^{4} - e ^{10} ) ) e ^{2 t}}{2 ( e ^{12} + 1 )} + \frac{4 e ^{4} - e ^{10}}{2 ( e ^{12} + 1 )} e^{- 2 t}

Schritte zur Lösung

y^{''} (t) - 4 y = 0

Solve linear ODE:

y = \frac{( e ^{2} ( e ^{12} + 1 ) + e ^{4} ( 4 e ^{4} - e ^{10} ) ) e ^{2 t}}{2 ( e ^{12} + 1 )} + \frac{4 e ^{4} - e ^{10}}{2 ( e ^{12} + 1 )} e^{- 2 t}

y = \frac{( e ^{2} ( e ^{12} + 1 ) + e ^{4} ( 4 e ^{4} - e ^{10} ) ) e ^{2 t}}{2 ( e ^{12} + 1 )} + \frac{4 e ^{4} - e ^{10}}{2 ( e ^{12} + 1 )} e^{- 2 t}

y^{^{'''}} + y^{^{''}} + y^{^{'}} - 39 y = 0

y^{^{''}} + 4 y + 68 y = 0, y (0) = 7

3 y^{^{''}} + 48 y = 0, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''''}} + 4 y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} - 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + \frac{df}{d x} = 0