What is the solution for y^{''}+8y^'+4y=t ?

The solution for y^{''}+8y^'+4y=t is y=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}+t/4-1/2

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y^{''}+8y^'+4y=t

y^{^{''}} + 8 y^{^{'}} + 4 y = t

y = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t} + \frac{t}{4} - \frac{1}{2}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 4 y = t

Solve linear ODE:

y = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t} + \frac{t}{4} - \frac{1}{2}

y = c_{1} e^{2 (3 - 2) t} + c_{2} e^{- 2 (2 + 3) t} + \frac{t}{4} - \frac{1}{2}

\frac{\partial}{\partial y} (sin (z))

4 x + 9, 144 - x, - 9, 144

\frac{2}{x ^{3}} - \frac{8}{x}

y = (\frac{f ( x )}{g ( x )})^{2}

\int x^{4} e^{2 x} d x

What is the solution for y^{''}+8y^'+4y=t ?
The solution for y^{''}+8y^'+4y=t is y=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}+t/4-1/2