e^yy^'=((ln(t))^2)/t ,y(1)=ln(15)

Lösung

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (15)

y = ln (\frac{ln ^{3} ( t )}{3} + 15)

Schritte zur Lösung

e^{y} y^{'} (t) = \frac{( ln ( t ) ) ^{2}}{t}

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + 15)

y = ln (\frac{ln ^{3} ( t )}{3} + 15)

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

\frac{d v}{d t} = 5 - 0.5 \cdot v

y^{^{'}} + 4 y = 3, y (0) = 0

(R + 1) d R - R (in (c)) d c = 0

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