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

Lösung

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

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{4 t ^{2}}{5} - \frac{26 t}{25} + \frac{12}{125}

Schritte zur Lösung

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{4 t ^{2}}{5} - \frac{26 t}{25} + \frac{12}{125}

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{4 t ^{2}}{5} - \frac{26 t}{25} + \frac{12}{125}

\frac{d}{d x} (\frac{3}{x ^{3}} - \frac{1}{x ^{4}})

\frac{d}{d x} (\frac{( x ^{2} - 3 x )}{( 2 x + 1 )})

3 y^{^{''}} + y = 0

\int_{0}^{1} 2 x \cdot 1 + 4 x^{2} d x

\int \frac{( 7 - x )}{( x ^{2} - 4 x + 4 )} d x