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

The solution for y^{''}+4y^'=-6x-3 is y=c_{1}+c_{2}e^{-4x}-(3x^2)/4-(3x)/8

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

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

y = c_{1} + c_{2} e^{- 4 x} - \frac{3 x ^{2}}{4} - \frac{3 x}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{3 x ^{2}}{4} - \frac{3 x}{8}

y = c_{1} + c_{2} e^{- 4 x} - \frac{3 x ^{2}}{4} - \frac{3 x}{8}

\frac{\partial}{\partial x} (- sin (x) cos (y))

n = 2 \sum \infty \frac{1}{( n ^{3} - n )}

\frac{d y}{d x} = \frac{1}{3} y (8 - y), y (0) = 1

2 x^{4} y^{^{'}} = k

\int (2 x^{3} + 6 x)^{2} d x

What is the solution for y^{''}+4y^'=-6x-3 ?
The solution for y^{''}+4y^'=-6x-3 is y=c_{1}+c_{2}e^{-4x}-(3x^2)/4-(3x)/8