What is the solution for y^'=-2y+t,y(0)=-5 ?

The solution for y^'=-2y+t,y(0)=-5 is y=(2t-1)/4-(19)/(4e^{2t)}

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y^'=-2y+t,y(0)=-5

y^{^{'}} = - 2 y + t, y (0) = - 5

y = \frac{2 t - 1}{4} - \frac{19}{4 e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{2 t - 1}{4} - \frac{19}{4 e ^{2 t}}

y = \frac{2 t - 1}{4} - \frac{19}{4 e ^{2 t}}

\int \frac{1}{y ^{2} - 3 y + 2} d y

\int (x^{2} + 3 x + 1) d x

\frac{d y}{d x} + (x^{2} - 1) y = - x y^{6}

\int 0.03 e^{- 0.03 x} d x

\frac{\partial}{\partial y} (arctan (yz))

What is the solution for y^'=-2y+t,y(0)=-5 ?
The solution for y^'=-2y+t,y(0)=-5 is y=(2t-1)/4-(19)/(4e^{2t)}