What is the solution for y^{''}-6y^'+8y=2x-1 ?

The solution for y^{''}-6y^'+8y=2x-1 is y=c_{1}e^{4x}+c_{2}e^{2x}+x/4+1/16

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y^{''}-6y^'+8y=2x-1

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 x - 1

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = 2 x - 1

Solve linear ODE:

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

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

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

(d^{2} + d + 1) y = e^{x}

3 \frac{d ^{2} x}{d t ^{2}} + 60 x = 2 cos (25 t)

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 = 0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 6 e x + 10 cos (x)

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