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

The solution for y^{''}+4y^'+3y=2x is y=c_{1}e^{-x}+c_{2}e^{-3x}+(2x)/3-8/9

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2 x}{3} - \frac{8}{9}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 2 x

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2 x}{3} - \frac{8}{9}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2 x}{3} - \frac{8}{9}

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

\int 1 - sin^{2} (x) d x

\frac{d}{d x} (16 + x)

x \to 0 lim (2^{- x})

\int_{1}^{2} x - 1 - x + 1 d x

What is the solution for y^{''}+4y^'+3y=2x ?
The solution for y^{''}+4y^'+3y=2x is y=c_{1}e^{-x}+c_{2}e^{-3x}+(2x)/3-8/9