What is the solution for y^{''}+3y^'+2y=\H(2) ?

The solution for y^{''}+3y^'+2y=\H(2) is y=c_{1}e^{-t}+c_{2}e^{-2t}+(Undefined)/2

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y^{''}+3y^'+2y=\H(2)

y^{^{''}} + 3 y^{^{'}} + 2 y = H (2)

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = H (2)

Solve linear ODE:

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

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

y^{^{''}} + n^{2} y = \frac{2}{n ^{2} π} ((- 1)^{n} - 1) cos (t)

y^{^{''}} - 10 y^{^{'}} + 25 y = 75 t + 295

(d^{2} - 2 d - 3) y = x^{3} e^{- 3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{1}{x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}+3y^'+2y=\H(2) ?
The solution for y^{''}+3y^'+2y=\H(2) is y=c_{1}e^{-t}+c_{2}e^{-2t}+(Undefined)/2