y^{''}-6y^'+9y=5t^2,y(0)=0,y^'(0)=1

Lösung

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{10}{27} e^{3 t} + \frac{37 e ^{3 t} t}{27} + \frac{5 t ^{2}}{9} + \frac{20 t}{27} + \frac{10}{27}

Schritte zur Lösung

y^{''} (t) - 6 y^{'} (t) + 9 y = 5 t^{2}

Solve linear ODE:

y = - \frac{10}{27} e^{3 t} + \frac{37 e ^{3 t} t}{27} + \frac{5 t ^{2}}{9} + \frac{20 t}{27} + \frac{10}{27}

y = - \frac{10}{27} e^{3 t} + \frac{37 e ^{3 t} t}{27} + \frac{5 t ^{2}}{9} + \frac{20 t}{27} + \frac{10}{27}

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 7

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 5

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = \frac{e ^{t}}{t}

y^{^{''}} - 2 y^{^{'}} - 15 y = 3 e^{5 t}

y^{^{'''}} + 4 y^{^{''}} - 5 y = x e^{x} + 8