What is the solution for y^{''}+vy^'=ce^{(-r^2)} ?

The solution for y^{''}+vy^'=ce^{(-r^2)} is y=c_{1}+c_{2}e^{-vt}+(ce^{-r^2}t)/v

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y^{''}+vy^'=ce^{(-r^2)}

y^{^{''}} + v y^{^{'}} = c e^{(- r^{2})}

y = c_{1} + c_{2} e^{- v t} + \frac{c e ^{- r^{2}} t}{v}

Solution steps

y^{''} (t) + v y^{'} (t) = c e^{(- r^{2})}

Solve linear ODE:

y = c_{1} + c_{2} e^{- v t} + \frac{c e ^{- r^{2}} t}{v}

y = c_{1} + c_{2} e^{- v t} + \frac{c e ^{- r^{2}} t}{v}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 e^{- 5 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (5 t)

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 x - e^{- 4 x} + sin (2 x)

y^{^{''}} + 25 y = 32 sin (3 x + 2)

What is the solution for y^{''}+vy^'=ce^{(-r^2)} ?
The solution for y^{''}+vy^'=ce^{(-r^2)} is y=c_{1}+c_{2}e^{-vt}+(ce^{-r^2}t)/v