What is the solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(19) ?

The solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(19) is y=ln((ln^3(t))/3+19)

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e^yy^'=((ln(t))^2)/t ,y(1)=ln(19)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (19)

y = ln (\frac{ln ^{3} ( t )}{3} + 19)

Solution steps

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

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + 19)

y = ln (\frac{ln ^{3} ( t )}{3} + 19)

y^{^{'}} = \frac{( y + 1 )}{( x - 1 )}

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

2 x y + 6 x + (x^{2} - 4) y^{^{'}} = 0, y (3) = 2

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}, y (0) = \frac{π}{4}

What is the solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(19) ?
The solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(19) is y=ln((ln^3(t))/3+19)