What is the solution for y^{''}-14y^'+49y=2xe^{7x} ?

The solution for y^{''}-14y^'+49y=2xe^{7x} is y=c_{1}e^{7x}+c_{2}xe^{7x}+(e^{7x}x^3)/3

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y^{''}-14y^'+49y=2xe^{7x}

y^{^{''}} - 14 y^{^{'}} + 49 y = 2 x e^{7 x}

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{e ^{7 x} x ^{3}}{3}

Solution steps

y^{''} (x) - 14 y^{'} (x) + 49 y = 2 x e^{7 x}

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{e ^{7 x} x ^{3}}{3}

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{e ^{7 x} x ^{3}}{3}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x, y (0) = - 1, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 = 0

- 5 y^{^{''}} - 2 y^{^{'}} + 24 = 0

y^{^{''}} - 3 y^{^{'}} = cos (5 x), y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{3} e^{3 t}

What is the solution for y^{''}-14y^'+49y=2xe^{7x} ?
The solution for y^{''}-14y^'+49y=2xe^{7x} is y=c_{1}e^{7x}+c_{2}xe^{7x}+(e^{7x}x^3)/3