What is the solution for t^2(dy)/(dt)-t=1+y+yt ?

The solution for t^2(dy)/(dt)-t=1+y+yt is y=e^{-1/t+c_{1}}t-1

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t^2(dy)/(dt)-t=1+y+yt

t^{2} \frac{d y}{d t} - t = 1 + y + y t

y = e^{- \frac{1}{t} + c_{1}} t - 1

Solution steps

t^{2} \frac{d y}{d t} - t = 1 + y + y t

Solve separable ODE:

y = e^{- \frac{1}{t} + c_{1}} t - 1

y = e^{- \frac{1}{t} + c_{1}} t - 1

y^{^{''}} = ey

\frac{d}{d x} (e^{x + 5})

y = ln (x - 1), y = x - 2, x = 3

\frac{100}{s ( s ^{2} + 2 s + 6 )}

f^{^{'}} (x) = sec^{2} (x)

What is the solution for t^2(dy)/(dt)-t=1+y+yt ?
The solution for t^2(dy)/(dt)-t=1+y+yt is y=e^{-1/t+c_{1}}t-1