(dy)/(dx)=e^{3x+2y},y(0)=0

Lösung

\frac{d y}{d x} = e^{3 x + 2 y}, y (0) = 0

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} + \frac{5}{3} )}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} + \frac{5}{3} )}{2}

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} + \frac{5}{3} )}{2}

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

\frac{d y}{d x} e^{x + y} - x = 0

x y^{^{'}} = 3 (y^{2} + y)

y^{^{'}} = x y, y (1) = 1

\frac{d v}{d t} + \frac{v}{3} = 4