xy^'+(1+x)=e^{-x}

Lösung

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

y = Ei (- x) - x - ln (x) + c_{1}

Schritte zur Lösung

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

Isoliere

y^{'} (x) : y^{'} (x) = \frac{e ^{- x} - x - 1}{x}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int \frac{e ^{- x} - x - 1}{x} d x

\int \frac{e ^{- x} - x - 1}{x} d x = Ei (- x) - x - ln (x) + c_{1}

y = Ei (- x) - x - ln (x) + c_{1}

\int sin (2 x) - 5 cos (4 x) d x

\int 10 x^{5} e^{x^{6}} d x

\frac{\partial}{\partial x} (ln (e^{x} + e^{- x}))

\frac{\partial}{\partial x} (\frac{1}{a} e^{- (\frac{x}{a})^{2}})

t an g e n t f (x) = 8 x^{2} + 3 x, at x = - 45