What is the solution for y^{''}+3y^'+2y=6 ?

The solution for y^{''}+3y^'+2y=6 is y=c_{1}e^{-t}+c_{2}e^{-2t}+3

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y^{''}+3y^'+2y=6

y^{^{''}} + 3 y^{^{'}} + 2 y = 6

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 3

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 6

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 3

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 3

x \to 6 lim (\frac{1}{x - 2})

f (x) = cos (x), 0, 2 π

y = - 2 - 7 x^{2}, (- 2, - 30)

\int \frac{3 cos ( x )}{sin ^{2} ( x ) + sin ( x )} d x

x^{2} - 35, 5 - 3 x

What is the solution for y^{''}+3y^'+2y=6 ?
The solution for y^{''}+3y^'+2y=6 is y=c_{1}e^{-t}+c_{2}e^{-2t}+3