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

Lösung

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

y = \frac{1}{4} e^{2 t} - \frac{13 e ^{2 t} t}{8} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{1}{4} e^{2 t} - \frac{13 e ^{2 t} t}{8} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

y = \frac{1}{4} e^{2 t} - \frac{13 e ^{2 t} t}{8} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

\frac{d y}{d x} = \frac{( y - x + 1 )}{y - x - 6}

y^{^{''''}} + 5 y^{^{''}} + 4 y = 0, y (0) = 0

y d x - 4 x d y = y^{6}

\int \frac{x ^{3}}{3} \frac{1}{x} d x

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