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

The solution for y^{''}-7y^'+12y=x is y=c_{1}e^{4x}+c_{2}e^{3x}+x/(12)+7/144

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

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

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{x}{12} + \frac{7}{144}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{x}{12} + \frac{7}{144}

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{x}{12} + \frac{7}{144}

3 \frac{d ^{2} x}{d t ^{2}} + 10 x = 2 sin (3 t)

y^{^{''}} + y = 6 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2} e^{3 x}

6 y^{^{''}} - y^{^{'}} - 7 y = - 7

y^{^{''}} - y^{^{'}} - 6 y = 58 cos (3 x)

What is the solution for y^{''}-7y^'+12y=x ?
The solution for y^{''}-7y^'+12y=x is y=c_{1}e^{4x}+c_{2}e^{3x}+x/(12)+7/144