-4y^{''}+16y=0,y(1)=2,y^'(2)=1

Lösung

- 4 y^{^{''}} + 16 y = 0, y (1) = 2, y^{^{'}} (2) = 1

y = e^{2 t - 4} (2 e^{2} - \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )}) + \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )} e^{- 2 t}

Schritte zur Lösung

- 4 y^{''} (t) + 16 y = 0

Solve linear ODE:

y = e^{2 t - 4} (2 e^{2} - \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )}) + \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )} e^{- 2 t}

y = e^{2 t - 4} (2 e^{2} - \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )}) + \frac{4 e ^{6} - e ^{4}}{2 ( e ^{4} + 1 )} e^{- 2 t}

d er i v a t i v e y = x^{3} (x - 1)^{2}

d er i v a t i v e f (x) = 6 sin^{3} (x)

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 1} d x

y^{^{''}} + 16 y^{^{'}} + 73 y = 0

y^{^{''}} - 8 y^{^{'}} + 4 y = 0