What is the solution for y^{''}+3y^'+2y=4 ?

The solution for y^{''}+3y^'+2y=4 is y=c_{1}e^{-t}+c_{2}e^{-2t}+2

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y^{''}+3y^'+2y=4

y^{^{''}} + 3 y^{^{'}} + 2 y = 4

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 4

Solve linear ODE:

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

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

\int csc^{3} (x) cot^{2} (x) d x

e^{θ}

\int \frac{( e ^{2 x} + 1 )}{e ^{x}} d x

y^{^{'}} = \frac{1}{y}, y (0) = 1

f (x) = 3 x^{2} + 1

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