y^{''}-8y^'+13y=0,y(0)=0,y^'(0)=7

Lösung

y^{^{''}} - 8 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = 7

y = \frac{7}{2 3} e^{(4 + 3) t} - \frac{7}{2 3} e^{(4 - 3) t}

Schritte zur Lösung

y^{''} (t) - 8 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = \frac{7}{2 3} e^{(4 + 3) t} - \frac{7}{2 3} e^{(4 - 3) t}

y = \frac{7}{2 3} e^{(4 + 3) t} - \frac{7}{2 3} e^{(4 - 3) t}

\int \frac{1}{( sec ( x ) + tan ( x ) ) ^{2}} d x

\frac{\partial}{\partial y} (3 x e^{- \frac{1}{y ^{3}}})

\int_{115}^{150} \frac{75}{2 x ^{2}} d x

d er i v a t i v e \frac{( x ^{2} - 3 x )}{x ^{2} + 7 x}

d er i v a t i v e y = 9 - x