y^'=e^{4x-5y}

Lösung

y^{^{'}} = e^{4 x - 5 y}

y = \frac{ln ( \frac{5}{4} e ^{4 x} + c _{1} )}{5}

Schritte zur Lösung

y^{'} (x) = e^{4 x - 5 y}

Solve separable ODE:

y = \frac{ln ( \frac{5}{4} e ^{4 x} + c _{1} )}{5}

y = \frac{ln ( \frac{5}{4} e ^{4 x} + c _{1} )}{5}

y^{^{'}} = \frac{6 e ^{2 x}}{e ^{y}}

y^{^{'}} = \frac{1 + x}{y}, y (0) = 5

y^{^{'}} sin (y) = 1

\frac{d x}{d t} = x^{2} + \frac{1}{16}, x (0) = 6

\frac{d v}{d x} = 9 + 16 v^{2}