What is the solution for y^{''}-14y^'+49y=35x+4 ?

The solution for y^{''}-14y^'+49y=35x+4 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(5x)/7+2/7

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y^{''}-14y^'+49y=35x+4

y^{^{''}} - 14 y^{^{'}} + 49 y = 35 x + 4

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{5 x}{7} + \frac{2}{7}

Solution steps

y^{''} (x) - 14 y^{'} (x) + 49 y = 35 x + 4

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{5 x}{7} + \frac{2}{7}

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{5 x}{7} + \frac{2}{7}

\frac{d ^{2} y}{d t ^{2}} + 4 y = sec^{2} (2 t)

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

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

3 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 2 = 0

y^{^{''}} + 3 y = cos (x)

What is the solution for y^{''}-14y^'+49y=35x+4 ?
The solution for y^{''}-14y^'+49y=35x+4 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(5x)/7+2/7