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

The solution for y^{''}+3y^'+2y=x is y=c_{1}e^{-x}+c_{2}e^{-2x}+x/2-3/4

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x}{2} - \frac{3}{4}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x}{2} - \frac{3}{4}

f (x) = 5 x^{2} - 8 x, at x = 1

x \to 2 lim (a x + b)

\frac{2 x ^{2}}{x ^{2} - 1}

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{2}})

\frac{6 x - 1}{2 x + 9}

What is the solution for y^{''}+3y^'+2y=x ?
The solution for y^{''}+3y^'+2y=x is y=c_{1}e^{-x}+c_{2}e^{-2x}+x/2-3/4