What is the solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(2) ?

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

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

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

y = ln (\frac{ln ^{3} ( t )}{3} + 2)

Solution steps

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

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + 2)

y = ln (\frac{ln ^{3} ( t )}{3} + 2)

\frac{d y}{d x} = \frac{y ( y + 1 )}{x}

y^{^{'}} - 5 y = 0, y (0) = 2

\frac{d y}{d t} = y + 1, y (0) = 0

\frac{d x}{d t} = 125 - \frac{x}{100}, x = 4000, t = 0

y^{^{'}} + 12 y = 1

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