2y^'-y=40

Lösung

2 y^{^{'}} - y = 40

y = e^{\frac{t + c _{1}}{2}} - 40

Schritte zur Lösung

2 y^{'} (t) - y = 40

Solve separable ODE:

y = e^{\frac{t + c _{1}}{2}} - 40

y = e^{\frac{t + c _{1}}{2}} - 40

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

y^{^{'}} = \frac{y}{t ( 1 + t )}

\frac{d p}{d t} = 4 pt, p (1) = 3

y - 4 y^{^{'}} + 11 y = 0

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