y^'=e^{3x-4y},y(0)=0

解答

y^{^{'}} = e^{3 x - 4 y}, y (0) = 0

y = \frac{ln ( \frac{4}{3} e ^{3 x} - \frac{1}{3} )}{4}

求解步骤

y^{'} (x) = e^{3 x - 4 y}

求解可分离方程:

y = \frac{ln ( \frac{4}{3} e ^{3 x} - \frac{1}{3} )}{4}

y = \frac{ln ( \frac{4}{3} e ^{3 x} - \frac{1}{3} )}{4}

\frac{d y}{d x} + 9 x^{8} y = 0, y = 6 e^{- x^{9}}

e^{y} y^{^{'}} = 1, y (0) = ln (x + c)

y^{^{'}} = x y^{2} sin (x), y (0) = 2

x^{2} y^{^{'}} - 2 y = 0

f^{^{'}} (x) = - 2 \cdot f (x)