2y^'-0.8y=100

Lösung

2 y^{^{'}} - 0.8 y = 100

y = \frac{5 e ^{\frac{2 t + c _{1}}{5}}}{4} - 125

Schritte zur Lösung

2 y^{'} (t) - 0.8 y = 100

Solve separable ODE:

y = \frac{5 e ^{\frac{2 t + c _{1}}{5}}}{4} - 125

y = \frac{5 e ^{\frac{2 t + c _{1}}{5}}}{4} - 125

y^{^{'}} = - 2 \cdot t \cdot y^{2}

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

y \frac{d y}{d x} = 5 x (y^{2} + 1)^{6}, y (8) = 2

\frac{d y}{d x} = \frac{4 y ^{\frac{3}{4}}}{x ^{2} + 1}

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