y^'=(t+1)/(ty+t)

Lösung

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

y = - 1 + 2 (t + ln (t) + c_{1}) + 1, y = - 2 (t + ln (t) + c_{1}) + 1 - 1

Schritte zur Lösung

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

Solve separable ODE:

y = - 1 + 2 (t + ln (t) + c_{1}) + 1, y = - 2 (t + ln (t) + c_{1}) + 1 - 1

y = - 1 + 2 (t + ln (t) + c_{1}) + 1, y = - 2 (t + ln (t) + c_{1}) + 1 - 1

\frac{d y}{d x} sin (π x) = y cos (π x)

y^{^{'}} = \frac{x ^{2} y ^{2}}{1 + x}

y^{^{'}} = \frac{x ^{2}}{y ^{3}}

2 sin (x) - y^{2} \frac{d y}{d x} = 0, y (0) = 1

x^{^{'}} = - 2 x + x