9x^2y^'=y^'+7xe^{-y}

解答

9 \cdot x^{2} \cdot y^{^{'}} = y^{^{'}} + 7 \cdot x \cdot e^{- y}

y = ln (\frac{7}{18} ln (9 x^{2} - 1) + c_{1})

求解步骤

9 \cdot x^{2} \cdot y^{'} (x) = y^{'} (x) + 7 \cdot x \cdot e^{- y}

求解可分离方程:

y = ln (\frac{7}{18} ln (9 x^{2} - 1) + c_{1})

y = ln (\frac{7}{18} ln (9 x^{2} - 1) + c_{1})

d er i v a t i v e f (x) = 6 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 7 x + 1

\frac{d}{d x} (cos^{2} (\frac{x}{2}))

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{- t}

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

\int_{0}^{1} 11 x x^{2} + 36 d x