4y^{''}-8y^'+7y=0

Lösung

4 y^{^{''}} - 8 y^{^{'}} + 7 y = 0

y = e^{t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Schritte zur Lösung

4 y^{''} (t) - 8 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y^{^{''}} + 64 y = sec^{2} (8 x)

\int cos (4 x + 1) d x

\int tan^{3} (1) d x

\frac{d}{d x} (a \cdot b^{x})

\int t^{2} d t