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

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

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

e^{y} y^{^{'}} + \frac{t}{y + 1} = \frac{2}{y + 1}

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

Solution steps

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

Solve separable ODE:

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

x \to \infty lim (\frac{1}{6 ^{x + 1}})

(6, 2), (- 3, 5)

\frac{2 x - 3}{4 x + 1}

\int (3 x - 1) x + 6 d x

ln (\frac{2 x + 1}{x ^{2} - 4})

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