What is the solution for y^{''}-14y^'+49y=21x+6 ?

The solution for y^{''}-14y^'+49y=21x+6 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(3x)/7+12/49

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y^{''}-14y^'+49y=21x+6

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 6

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{3 x}{7} + \frac{12}{49}

Solution steps

y^{''} (x) - 14 y^{'} (x) + 49 y = 21 x + 6

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{3 x}{7} + \frac{12}{49}

y = c_{1} e^{7 x} + c_{2} x e^{7 x} + \frac{3 x}{7} + \frac{12}{49}

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

D (t) = 30 e^{- 0.1 (3)}

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}

y^{^{''}} - 8 y^{^{'}} + 16 = 0, y (0) = 4, y (1) = 5

4 y^{^{''}} + 6 y^{^{'}} + 25 y = 4 cos (2 t)

What is the solution for y^{''}-14y^'+49y=21x+6 ?
The solution for y^{''}-14y^'+49y=21x+6 is y=c_{1}e^{7x}+c_{2}xe^{7x}+(3x)/7+12/49