What is the solution for y^{''}-4y^'+4y=14x+5 ?

The solution for y^{''}-4y^'+4y=14x+5 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(7x)/2+19/4

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y^{''}-4y^'+4y=14x+5

y^{^{''}} - 4 y^{^{'}} + 4 y = 14 x + 5

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{7 x}{2} + \frac{19}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 14 x + 5

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{7 x}{2} + \frac{19}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{7 x}{2} + \frac{19}{4}

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{\frac{- 10 x}{x ^{2}}}

d (r) = \frac{( ln ( 2 ) )}{ln ( 1 + 0.04 )}

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

x^{^{''}} - (π^{2}) x = - \frac{1}{π}

2 y^{^{''}} - 4 y^{^{'}} - 6 y = 16 e^{- x}

What is the solution for y^{''}-4y^'+4y=14x+5 ?
The solution for y^{''}-4y^'+4y=14x+5 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(7x)/2+19/4