xy^'+4y=((2e^{-4x}))/(x^3)

Lösung

x y^{^{'}} + 4 y = \frac{( 2 e ^{- 4 x} )}{x ^{3}}

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

Schritte zur Lösung

x y^{'} (x) + 4 y = \frac{( 2 e ^{- 4 x} )}{x ^{3}}

Solve linear ODE:

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

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

3 x^{2} y + \frac{d y}{d x} = 4 x^{2}

\int 5 x + \frac{1}{5 x} d x

n = 0 \sum \infty 3.2 5^{n}

t an g e n t f (x) = \frac{1}{3 + 2 x}, at x = 1