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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 2 e^{u} d u

\frac{( 3 y ^{2} - x ^{2} )}{y ^{5}} y^{^{'}} + \frac{x}{2 y ^{4}} = 0

f (x) = - 5.7 + 3.57 ln (x)

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

\frac{d}{d x} (\frac{3 x - 1}{x + 1})

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