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

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

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t^2y^'-t=1+y+ty,y(1)=1

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

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

Solution steps

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

Solve separable ODE:

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

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

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

4 y \frac{d y}{d x} = 6 x^{2}

y^{3} \frac{d y}{d x} = 5 x^{2} y^{4} - 20 x^{2}

y \frac{d y}{d x} = x^{2}, y (0) = 1

\frac{d}{d x} (f (x)^{- 1}) x^{2} - 8 x - 9 x = - 21

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