y^{''}-35y^'+9y=60t,y(0)=-1,y^'(0)=2

Lösung

y^{^{''}} - 35 y^{^{'}} + 9 y = 60 t, y (0) = - 1, y^{^{'}} (0) = 2

y = (\frac{25193 + 727 1189}{54 1189} - \frac{727}{27}) e^{\frac{( 35 + 1189 ) t}{2}} - \frac{25193 + 727 1189}{54 1189} e^{\frac{( 35 - 1189 ) t}{2}} + \frac{20 t}{3} + \frac{700}{27}

Schritte zur Lösung

y^{''} (t) - 35 y^{'} (t) + 9 y = 60 t

Solve linear ODE:

y = (\frac{25193 + 727 1189}{54 1189} - \frac{727}{27}) e^{\frac{( 35 + 1189 ) t}{2}} - \frac{25193 + 727 1189}{54 1189} e^{\frac{( 35 - 1189 ) t}{2}} + \frac{20 t}{3} + \frac{700}{27}

y = (\frac{25193 + 727 1189}{54 1189} - \frac{727}{27}) e^{\frac{( 35 + 1189 ) t}{2}} - \frac{25193 + 727 1189}{54 1189} e^{\frac{( 35 - 1189 ) t}{2}} + \frac{20 t}{3} + \frac{700}{27}

y^{^{''}} + 2 y^{^{'}} + 5 y = 52 cos (3 x)

y^{^{''}} - 5 y^{^{'}} - 24 y = 5 + 2 t

y^{^{''}} + 36 y = 33 cos (5 x)

d (x) = 2 x - 4

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 sin (2 x) - 3 e^{2} x