What is the solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(10) ?

The solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(10) is y=ln((ln^3(t))/3+10)

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e^yy^'=((ln(t))^2)/t ,y(1)=ln(10)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (10)

y = ln (\frac{ln ^{3} ( t )}{3} + 10)

Solution steps

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

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + 10)

y = ln (\frac{ln ^{3} ( t )}{3} + 10)

(3 x cos (3 x) + sin (3 x) - 3) d x + (2 y + 5) d y = 0

sin (2 x) d y - e^{y} d x = 0

y^{^{'}} = 9 y^{- 2}, y (0) = 3

\frac{d x}{d y} = (6 x - 1) e^{y}

\frac{d y}{d t} = - 5 y, y (0) = 8

What is the solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(10) ?
The solution for e^yy^'=((ln(t))^2)/t ,y(1)=ln(10) is y=ln((ln^3(t))/3+10)