What is the solution for y^{''}-6y^'+8y=x^2-5x ?

The solution for y^{''}-6y^'+8y=x^2-5x is y=c_{1}e^{4x}+c_{2}e^{2x}+(x^2)/8-(7x)/(16)-23/64

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y^{''}-6y^'+8y=x^2-5x

y^{^{''}} - 6 y^{^{'}} + 8 y = x^{2} - 5 x

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x ^{2}}{8} - \frac{7 x}{16} - \frac{23}{64}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = x^{2} - 5 x

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x ^{2}}{8} - \frac{7 x}{16} - \frac{23}{64}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x ^{2}}{8} - \frac{7 x}{16} - \frac{23}{64}

y^{^{''}} - y^{^{'}} + y = 17 sin (t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 2 e^{- x}

3 x^{^{''}} + 3 x = 5 cos (4 t)

y^{^{''}} + \frac{1}{8} y^{^{'}} + y = 3 \cdot cos (x)

y^{^{''}} + 5 y^{^{'}} - 6 y = 21 e^{t}

What is the solution for y^{''}-6y^'+8y=x^2-5x ?
The solution for y^{''}-6y^'+8y=x^2-5x is y=c_{1}e^{4x}+c_{2}e^{2x}+(x^2)/8-(7x)/(16)-23/64