(y^'+1)/x =(e^x)/(x^2)

解答

\frac{y ^{^{'}} + 1}{x} = \frac{e ^{x}}{x ^{2}}

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

求解步骤

\frac{y ^{'} ( x ) + 1}{x} = \frac{e ^{x}}{x ^{2}}

整理

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

若

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

则

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

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

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

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

f^{^{'}} (x) = x^{\frac{1}{x}}

18 x + y^{^{'}} = 20, - 3 x y (1) = 2

2 y^{^{'}} = \frac{( 1 - sin ( x ) ) ^{- 1}}{2}

f^{^{'}} (t) = e^{- t} \cdot (8 - 8 t)

y^{^{'''}} = x^{4}