y^{''}-4y^'+5y=e^2x,y(0)=1,y^'(0)=3

Lösung

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2} x, y (0) = 1, y^{^{'}} (0) = 3

y = e^{2 x} ((1 - \frac{4 e ^{2}}{25}) cos (x) + (\frac{3 e ^{2}}{25} + 1) sin (x)) + \frac{e ^{2} x}{5} + \frac{4 e ^{2}}{25}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 5 y = e^{2} x

Solve linear ODE:

y = e^{2 x} ((1 - \frac{4 e ^{2}}{25}) cos (x) + (\frac{3 e ^{2}}{25} + 1) sin (x)) + \frac{e ^{2} x}{5} + \frac{4 e ^{2}}{25}

y = e^{2 x} ((1 - \frac{4 e ^{2}}{25}) cos (x) + (\frac{3 e ^{2}}{25} + 1) sin (x)) + \frac{e ^{2} x}{5} + \frac{4 e ^{2}}{25}

8 y^{^{''}} - y^{^{'}} + y + x^{2} = 0

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 8 e^{3 x}

\frac{d ^{2} x}{d t ^{2}} + 81 x = 8 sin (9 t)

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

x^{^{''}} - 9 x = 2 t^{2} - cos (t)