(dQ)/(dt)=4Q^2ln(t),Q(1)=-1

Lösung

\frac{d Q}{d t} = 4 Q^{2} ln (t), Q (1) = - 1

Q = - \frac{1}{4 t ln ( t ) - 4 t + 5}

Schritte zur Lösung

\frac{d Q}{d t} = 4 Q^{2} ln (t)

Solve separable ODE:

Q = - \frac{1}{4 t ln ( t ) - 4 t + 5}

Q = - \frac{1}{4 t ln ( t ) - 4 t + 5}

\frac{d p}{d t} = t^{2} p - p + t^{2} - 1

x d y - y d x = 2 d x

\frac{d y}{d x} = 6 x y^{8}

y y^{^{'}} = sin (x) e^{x + 2 y}

y^{^{'}} + 0.5 y = y^{3}