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

a re a f (x) = cos (x), 0, 2 π

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

n = 3 \sum \infty \frac{n}{e ^{3 n}}

\frac{d}{d x} (\frac{x ^{3}}{x ^{2} - 4})

a re a y = x^{2} - 7 x + 6, x = 2, x = 6

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