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

Lösung

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

y = e^{\frac{x}{2}} (- \frac{1}{2} cos (\frac{7 x}{2}) - \frac{1}{2 7} sin (\frac{7 x}{2})) + \frac{1}{2} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{x}{2}} (- \frac{1}{2} cos (\frac{7 x}{2}) - \frac{1}{2 7} sin (\frac{7 x}{2})) + \frac{1}{2} e^{x}

y = e^{\frac{x}{2}} (- \frac{1}{2} cos (\frac{7 x}{2}) - \frac{1}{2 7} sin (\frac{7 x}{2})) + \frac{1}{2} e^{x}

\int - \frac{1}{2 x ^{3}} d x

\int \frac{y ^{2} - 1}{y \frac{3}{2}} d y

\int (2 x - 1) sin (3 x) d x

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

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 4 e^{- x}