What is the solution for y^{''}-3y^'=-6x+4 ?

The solution for y^{''}-3y^'=-6x+4 is y=c_{1}e^{3x}+c_{2}+x^2-(2x)/3

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y^{''}-3y^'=-6x+4

y^{^{''}} - 3 y^{^{'}} = - 6 x + 4

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

Solution steps

y^{''} (x) - 3 y^{'} (x) = - 6 x + 4

Solve linear ODE:

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

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

x^{2} - 4 x, x = 0

\frac{d}{d x} (\frac{A + B x + C x ^{2}}{x ^{2}})

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

\frac{\partial}{\partial x} (4 e^{x} - 2 x e^{y})

ln (u)

What is the solution for y^{''}-3y^'=-6x+4 ?
The solution for y^{''}-3y^'=-6x+4 is y=c_{1}e^{3x}+c_{2}+x^2-(2x)/3