What is the solution for y^{''}-7y^'+12=x ?

The solution for y^{''}-7y^'+12=x is y=c_{1}e^{7x}+c_{2}-(x^2)/(14)+(83x)/(49)

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y^{''}-7y^'+12=x

y^{^{''}} - 7 y^{^{'}} + 12 = x

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{2}}{14} + \frac{83 x}{49}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 12 = x

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{2}}{14} + \frac{83 x}{49}

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{2}}{14} + \frac{83 x}{49}

y^{^{''}} + y^{^{'}} = \frac{1}{sin ( x )}

y^{^{''}} + 16 y = 20 sin (4 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 6 x e^{- 3 x} + 27 x + 27

y^{^{''}} + 3 y^{^{'}} - 10 y = - 42 t + 20 t^{2}

What is the solution for y^{''}-7y^'+12=x ?
The solution for y^{''}-7y^'+12=x is y=c_{1}e^{7x}+c_{2}-(x^2)/(14)+(83x)/(49)