(x-1)y^'+xy=sin(x)

Lösung

(x - 1) y^{^{'}} + x y = sin (x)

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

Schritte zur Lösung

(x - 1) y^{'} (x) + x y = sin (x)

Solve linear ODE:

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

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

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

y^{^{'}} + y + e^{- x} = 0

y^{^{'}} = 2 x y + 1

\frac{y ^{^{'}}}{t + 3 y} = 3

1 + x y = x y^{^{'}}