e^yy^'=((ln(t))^2)/t ,y(1)=ln(4)

Lösung

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (4)

y = ln (\frac{ln ^{3} ( t )}{3} + 4)

Schritte zur Lösung

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

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + 4)

y = ln (\frac{ln ^{3} ( t )}{3} + 4)

(y^{2} - 1) d x + x d y = 0

y^{3} \frac{d y}{d x} = 3 x^{4} y^{4} - 12 x^{4}

y^{^{'}} - y \cdot tan (x) = 0

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

2 x y = - (x^{2} - 1) y^{^{'}}