What is the solution for e^yy^'=(((ln(t))^2))/t ?

The solution for e^yy^'=(((ln(t))^2))/t is y=ln((ln^3(t))/3+c_{1})

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e^yy^'=(((ln(t))^2))/t

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

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

\frac{d y}{d u} + 3 u^{2} y = 5 u^{2}

\frac{d Q}{d t} = \frac{50 - Q}{10}

y^{2} (x d y - y d x) = 9 x^{2} (y d x - x d y)

\frac{d z}{d t} = 2 t e^{2 z}

\frac{d x}{d y} + 3 x^{2} y = 5 x^{2}

What is the solution for e^yy^'=(((ln(t))^2))/t ?
The solution for e^yy^'=(((ln(t))^2))/t is y=ln((ln^3(t))/3+c_{1})