y^{''}-2y^'+10y=x^2*e^x

Lösung

y^{^{''}} - 2 y^{^{'}} + 10 y = x^{2} \cdot e^{x}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x ^{2}}{9} - \frac{2}{81} e^{x}

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 10 y = x^{2} \cdot e^{x}

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x ^{2}}{9} - \frac{2}{81} e^{x}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{e ^{x} x ^{2}}{9} - \frac{2}{81} e^{x}

t an g e n t f (x) = (5 x - x^{2}) (5 - x - x^{2}), at x = 1

t an g e n t f (x) = x - 3

l a pl a ce t r an s f or m \frac{1}{2} cos (2 t)

\int \frac{1}{u a ^{2} - u ^{2}} d u

\frac{\partial}{\partial z} (cos (x yz))