y^{''}-4y^'+4y=5t^2,y(1)=0

Lösung

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}, y (1) = 0

y = \frac{e ^{2 t - 2} ( - 8 e ^{2} c _{2} - 45 )}{8} + c_{2} e^{2 t} t + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) + 4 y = 5 t^{2}

Solve linear ODE:

y = \frac{e ^{2 t - 2} ( - 8 e ^{2} c _{2} - 45 )}{8} + c_{2} e^{2 t} t + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

y = \frac{e ^{2 t - 2} ( - 8 e ^{2} c _{2} - 45 )}{8} + c_{2} e^{2 t} t + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

\frac{d q}{20 - q} = \frac{1}{16} d T

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

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{2 x} + 1

4 y^{^{''}} - y = e^{\frac{x}{2}} x

\frac{d ^{2} y}{d x ^{2}} + y = 3 sin (2 x) + x cos (2 x)