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))

\int_{0}^{7} 2 e^{x} d x

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

\frac{0.32}{( 15.5 \cdot s + 1 ) \cdot s}

\frac{1}{( s ^{2} + 167 s + 12874 )}

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