What is the solution for t^2(dy)/(dt)-t=1+y+ty,y(1)=8 ?

The solution for t^2(dy)/(dt)-t=1+y+ty,y(1)=8 is y=9e^{-1/t+1}t-1

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t^2(dy)/(dt)-t=1+y+ty,y(1)=8

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 8

y = 9 e^{- \frac{1}{t} + 1} t - 1

Solution steps

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

Solve separable ODE:

y = 9 e^{- \frac{1}{t} + 1} t - 1

y = 9 e^{- \frac{1}{t} + 1} t - 1

y = \frac{ln ( 6 x )}{6 x}

\frac{d ^{2}}{d x ^{2}} (11 x)

n = 1 \sum \infty \frac{6}{n ( n + 2 )}

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

\int_{0}^{1} (x^{2} + 2) e^{- x} d x

What is the solution for t^2(dy)/(dt)-t=1+y+ty,y(1)=8 ?
The solution for t^2(dy)/(dt)-t=1+y+ty,y(1)=8 is y=9e^{-1/t+1}t-1