What is the solution for y^{''}-8y^'+16y=24x+2 ?

The solution for y^{''}-8y^'+16y=24x+2 is y=c_{1}e^{4x}+c_{2}xe^{4x}+(3x)/2+7/8

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y^{''}-8y^'+16y=24x+2

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3 x}{2} + \frac{7}{8}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 24 x + 2

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3 x}{2} + \frac{7}{8}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{3 x}{2} + \frac{7}{8}

\frac{d ^{2} y}{d x ^{2}} + y = - 16 cos (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{1}{1 + e ^{- x}}

x^{^{''}} + 8 x^{^{'}} + 16 x = 2 sin (4 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 7 e^{3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 x - 5 + 6 x e^{3 x}

What is the solution for y^{''}-8y^'+16y=24x+2 ?
The solution for y^{''}-8y^'+16y=24x+2 is y=c_{1}e^{4x}+c_{2}xe^{4x}+(3x)/2+7/8