y^{''}-4y^'=e^x

Lösung

y^{^{''}} - 4 y^{^{'}} = e^{x}

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) = e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

4 \cdot y^{^{''}} - 11 \cdot y^{^{'}} - 3 y = 5 \cdot e^{3 x}

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t), x (0) = 3

y^{^{''}} - 12 y^{^{'}} + 36 y = 30 x + 4

y^{^{''}} - y^{^{'}} = 5

y^{^{''}} - y^{^{'}} = 2