What is the solution for y^{''}-4y^'-x+2=0 ?

The solution for y^{''}-4y^'-x+2=0 is y=c_{1}e^{4x}+c_{2}-(x^2)/8+(7x)/(16)

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y^{''}-4y^'-x+2=0

y^{^{''}} - 4 y^{^{'}} - x + 2 = 0

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{8} + \frac{7 x}{16}

Solution steps

y^{''} (x) - 4 y^{'} (x) - x + 2 = 0

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{8} + \frac{7 x}{16}

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{8} + \frac{7 x}{16}

\frac{d}{d x} (x^{3} + 2 x^{2} - 4 x - 8)

x \to 4 lim (3 e^{1.7 x} + 1)

y^{^{''}} - 4 y^{^{'}} + 4 y = 8 x^{2} + e^{2 x}

y = 3 x^{3} + 45 x^{2} + 81 x

\int \frac{1}{x x + 4} d x

What is the solution for y^{''}-4y^'-x+2=0 ?
The solution for y^{''}-4y^'-x+2=0 is y=c_{1}e^{4x}+c_{2}-(x^2)/8+(7x)/(16)