-y^'+y=10

Lösung

- y^{^{'}} + y = 10

y = - e^{t + c_{1}} + 10

Schritte zur Lösung

- y^{'} (t) + y = 10

Solve separable ODE:

y = - e^{t + c_{1}} + 10

y = - e^{t + c_{1}} + 10

\frac{d y}{d x} = - \frac{3}{9} x^{5} y^{8}

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (15)

\frac{d y}{d x} = - \frac{y}{a}

\frac{d v}{d t} = 5 - 0.5 \cdot v

y^{^{'}} + 4 y = 3, y (0) = 0