y^2+y=t*y^',y(1)=-2

Lösung

y^{2} + y = t \cdot y^{^{'}}, y (1) = - 2

y = \frac{2 t}{1 - 2 t}

Schritte zur Lösung

y^{2} + y = t \cdot y^{'} (t)

Solve Bernoulli ODE:

y = \frac{2 t}{1 - 2 t}

y = \frac{2 t}{1 - 2 t}

y^{^{'}} = \frac{y}{3 x - 1}

\frac{d x}{d t} = 1.8 - \frac{5 x}{100}

\frac{d v}{d t} = 10 - 0.2 v

2 x \cdot y^{^{'}} + y = \frac{4}{y ^{2}}

\frac{d y}{d t} + 0.9 t y = 4 t, y (0) = 6