x^'=(1+t)/(2tx)

Lösung

x^{^{'}} = \frac{1 + t}{2 t x}

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

Schritte zur Lösung

x^{'} (t) = \frac{1 + t}{2 t x}

Solve separable ODE:

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

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

\frac{d y}{d t} + 1 + t^{2} y = 0

y^{^{'}} = - 3 y e^{t}, y (0) = 1

2 x y^{2} - x^{2} y + (2 y - x) y^{^{'}} = 0

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

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