What is the solution for y^{''}+3y^'=4x-5 ?

The solution for y^{''}+3y^'=4x-5 is y=c_{1}+c_{2}e^{-3x}+(2x^2)/3-(19x)/9

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y^{''}+3y^'=4x-5

y^{^{''}} + 3 y^{^{'}} = 4 x - 5

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{2}}{3} - \frac{19 x}{9}

Solution steps

y^{''} (x) + 3 y^{'} (x) = 4 x - 5

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{2}}{3} - \frac{19 x}{9}

y = c_{1} + c_{2} e^{- 3 x} + \frac{2 x ^{2}}{3} - \frac{19 x}{9}

y^{^{''}} + 4 y^{^{'}} + 4 = 0, y^{^{'}} (0) = 1

x \to \infty lim (- \frac{1}{x})

f (x) = cos (x), - \frac{π}{2}, \frac{π}{2}

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

\frac{d y}{d x} + \frac{y}{x} = x y^{2}

What is the solution for y^{''}+3y^'=4x-5 ?
The solution for y^{''}+3y^'=4x-5 is y=c_{1}+c_{2}e^{-3x}+(2x^2)/3-(19x)/9