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

The solution for y^{''}-6y^'+8y=x is y=c_{1}e^{4x}+c_{2}e^{2x}+x/8+3/32

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

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

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{8} + \frac{3}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{8} + \frac{3}{32}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{8} + \frac{3}{32}

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

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{2 t}

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

y^{^{''''}} + y^{^{''}} = x^{2} + x

y^{^{''}} - y^{^{'}} - 6 y = e^{3 x} + 30

What is the solution for y^{''}-6y^'+8y=x ?
The solution for y^{''}-6y^'+8y=x is y=c_{1}e^{4x}+c_{2}e^{2x}+x/8+3/32