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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

d y + y tan (x) d x = 0

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

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

(2 + t) y^{^{'}} + y = - 1

\frac{d y}{d x} 8 x^{2} - y^{2} = 6