What is the solution for y^{''}+3y^'+y=1 ?

The solution for y^{''}+3y^'+y=1 is y=c_{1}e^{((-3+sqrt(5))t)/2}+c_{2}e^{((-3-sqrt(5))t)/2}+1

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y^{''}+3y^'+y=1

y^{^{''}} + 3 y^{^{'}} + y = 1

y = c_{1} e^{\frac{( - 3 + 5 ) t}{2}} + c_{2} e^{\frac{( - 3 - 5 ) t}{2}} + 1

Solution steps

y^{''} (t) + 3 y^{'} (t) + y = 1

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 5 ) t}{2}} + c_{2} e^{\frac{( - 3 - 5 ) t}{2}} + 1

y = c_{1} e^{\frac{( - 3 + 5 ) t}{2}} + c_{2} e^{\frac{( - 3 - 5 ) t}{2}} + 1

\frac{d}{d x} ((x^{2} + 2 x - 1) \cdot (2 x^{2} - 3 x + 6))

\int 2 e^{- x} d x

h \to 0 lim (\frac{1}{h + 1})

x^{2} + 2 x y - y^{2} + x = 6, (2, 4)

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

What is the solution for y^{''}+3y^'+y=1 ?
The solution for y^{''}+3y^'+y=1 is y=c_{1}e^{((-3+sqrt(5))t)/2}+c_{2}e^{((-3-sqrt(5))t)/2}+1