What is the solution for y^'+1/t y=e^{-t^2},y(1)=12 ?

The solution for y^'+1/t y=e^{-t^2},y(1)=12 is y=-(e^{-t^2})/(2t)+(24e+1)/(2et)

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y^'+1/t y=e^{-t^2},y(1)=12

y^{^{'}} + \frac{1}{t} y = e^{- t^{2}}, y (1) = 12

y = - \frac{e ^{- t^{2}}}{2 t} + \frac{24 e + 1}{2 e t}

Solution steps

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

Solve linear ODE:

y = - \frac{e ^{- t^{2}}}{2 t} + \frac{24 e + 1}{2 e t}

y = - \frac{e ^{- t^{2}}}{2 t} + \frac{24 e + 1}{2 e t}

\frac{d}{d x} (lo g_{2} ((5 x - 1)^{3}))

\frac{d}{d x} (\frac{x}{x + \frac{b}{x}})

x \to 3 + lim (a^{2} x - 12 a)

(2 x - 1) d x + (5 y + 4) d y = 0

y = x^{3}, y = 0, x = - 3, x = 3

What is the solution for y^'+1/t y=e^{-t^2},y(1)=12 ?
The solution for y^'+1/t y=e^{-t^2},y(1)=12 is y=-(e^{-t^2})/(2t)+(24e+1)/(2et)