3y^'+2y^{''}=-10cos(3.14*t/(0.2))

Lösung

3 y^{^{'}} + 2 y^{^{''}} = - 10 cos (3.14 \cdot \frac{t}{0.2})

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - 0.00192 \dots sin (15.7 t) + 0.02010 \dots cos (15.7 t)

Schritte zur Lösung

3 y^{'} (t) + 2 y^{''} (t) = - 10 cos (3.14 \cdot \frac{t}{0.2})

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - 0.00192 \dots sin (15.7 t) + 0.02010 \dots cos (15.7 t)

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - 0.00192 \dots sin (15.7 t) + 0.02010 \dots cos (15.7 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = 7 + 4 x

y^{^{''}} - 3 y + 2 y = sin (e^{- x})

y^{^{'''}} + 16 y^{^{'}} = t

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{x}{e ^{x}} + \frac{x ^{2}}{e ^{2 x}}

y^{^{''''}} - 5 y^{^{''}} + 4 = 0