y^{''}-3y^'+y=2e^{3t},y(0)=0,y^'(0)=4

Lösung

y^{^{''}} - 3 y^{^{'}} + y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 4

y = (\frac{1 + 5}{5} - 2) e^{\frac{( 3 + 5 ) t}{2}} - \frac{1 + 5}{5} e^{\frac{( 3 - 5 ) t}{2}} + 2 e^{3 t}

Schritte zur Lösung

y^{''} (t) - 3 y^{'} (t) + y = 2 e^{3 t}

Solve linear ODE:

y = (\frac{1 + 5}{5} - 2) e^{\frac{( 3 + 5 ) t}{2}} - \frac{1 + 5}{5} e^{\frac{( 3 - 5 ) t}{2}} + 2 e^{3 t}

y = (\frac{1 + 5}{5} - 2) e^{\frac{( 3 + 5 ) t}{2}} - \frac{1 + 5}{5} e^{\frac{( 3 - 5 ) t}{2}} + 2 e^{3 t}

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{2 t}

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 12 y = 4

\frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 32 y = 32 H (t)

so l v e f or d, d (t) = - t^{2} + 5 t

y^{^{''}} + 2 y^{^{'}} + 2 y = 2 (x + 1)^{2}