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

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

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y^'-((t+1)y)/(t^2)= 1/t+1/(t^2)

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} = - (- \frac{x ^{2}}{3 y})

\frac{d y}{d x} = 2 x^{3} y

(x^{2} + 1) \frac{d y}{d x} + 3 x y = 3 x

\frac{d y}{d x} y = 2 x^{2} - 13 x + 6

\frac{d y}{d x} = \frac{( x + x y ^{2} )}{4 y}

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