What is the solution for y^{''}-14y^'+49y=28x+6 ?

The solution for y^{''}-14y^'+49y=28x+6 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(4x)/7+2/7

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y^{''}-14y^'+49y=28x+6

y^{^{''}} - 14 y^{^{'}} + 49 y = 28 x + 6

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

Solution steps

y^{''} (x) - 14 y^{'} (x) + 49 y = 28 x + 6

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 2 e^{6 x}

y^{^{''}} - y^{^{'}} - 6 y = 53 cos (3 x)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 t}}{t}

y^{^{''}} - 5 y^{^{'}} + 7 y = 3 e^{5 t}

What is the solution for y^{''}-14y^'+49y=28x+6 ?
The solution for y^{''}-14y^'+49y=28x+6 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(4x)/7+2/7