What is the solution for y^{''}+3y^'+2y=cos(x) ?

The solution for y^{''}+3y^'+2y=cos(x) is y=c_{1}e^{-x}+c_{2}e^{-2x}+3/10 sin(x)+1/10 cos(x)

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y^{''}+3y^'+2y=cos(x)

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = cos (x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

x^{^{''}} + 3 x^{^{'}} + 33 x = 9 cos (5 t)

y^{^{''}} + 4 y^{^{'}} + 3 y - 3 x = 0

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 3 y = x^{2} - 1

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 9 t

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (6 t)

What is the solution for y^{''}+3y^'+2y=cos(x) ?
The solution for y^{''}+3y^'+2y=cos(x) is y=c_{1}e^{-x}+c_{2}e^{-2x}+3/10 sin(x)+1/10 cos(x)