What is the solution for y^{''}-14y^'+49y=42x+5 ?

The solution for y^{''}-14y^'+49y=42x+5 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(6x)/7+17/49

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y^{''}-14y^'+49y=42x+5

y^{^{''}} - 14 y^{^{'}} + 49 y = 42 x + 5

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{6 x}{7} + \frac{17}{49}

Solution steps

y^{''} (x) - 14 y^{'} (x) + 49 y = 42 x + 5

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{6 x}{7} + \frac{17}{49}

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{6 x}{7} + \frac{17}{49}

d (x) = 5 (x - 9)^{2} - 32

y^{^{''}} + 4 y = 4 + 2 x

2 x^{^{''}} + 32 x = 68 e^{- 2 t} cos (4 t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = sin (x) - cos (x)

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

What is the solution for y^{''}-14y^'+49y=42x+5 ?
The solution for y^{''}-14y^'+49y=42x+5 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(6x)/7+17/49