What is the solution for y^{''}+4y^'=4x+1 ?

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

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y^{''}+4y^'=4x+1

y^{^{''}} + 4 y^{^{'}} = 4 x + 1

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

Solution steps

y^{''} (x) + 4 y^{'} (x) = 4 x + 1

Solve linear ODE:

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

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

(d^{2} - 4 d + 3) y = e^{- 2 t}

y^{^{''}} + y^{^{'}} - 12 = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 1296 t e^{- t}

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

y^{^{''}} + 9 y = cos (3 t) + sin (3 t)

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