What is the solution for y^'+2yt=0,y(1)=((e+3))/2 ?

The solution for y^'+2yt=0,y(1)=((e+3))/2 is y=(e+3)/2 e^{-t^2+1}

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y^'+2yt=0,y(1)=((e+3))/2

y^{^{'}} + 2 y t = 0, y (1) = \frac{( e + 3 )}{2}

y = \frac{e + 3}{2} e^{- t^{2} + 1}

Solution steps

y^{'} (t) + 2 y t = 0

Solve separable ODE:

y = \frac{e + 3}{2} e^{- t^{2} + 1}

y = \frac{e + 3}{2} e^{- t^{2} + 1}

\frac{d y}{d x} = 14 e^{5 x - 4 y}

(4 x - 8) d x = (6 y - 10) d y

e^{x} 3 y y^{^{'}} = e^{- y} + e^{- 4 x - y}

y^{^{'}} = \frac{9 x}{y + x ^{2} y}, y (0) = - 7

\frac{d y}{d x} = - \frac{4}{7} x^{3} y^{3}

What is the solution for y^'+2yt=0,y(1)=((e+3))/2 ?
The solution for y^'+2yt=0,y(1)=((e+3))/2 is y=(e+3)/2 e^{-t^2+1}