What is the solution for y^{''}-5y^'+6y=9x+2e^x ?

The solution for y^{''}-5y^'+6y=9x+2e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+(3x)/2+5/4+e^x

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y^{''}-5y^'+6y=9x+2e^x

y^{^{''}} - 5 y^{^{'}} + 6 y = 9 x + 2 e^{x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{3 x}{2} + \frac{5}{4} + e^{x}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 9 x + 2 e^{x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{3 x}{2} + \frac{5}{4} + e^{x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{3 x}{2} + \frac{5}{4} + e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 26 x + 47

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

y^{^{''}} - y^{^{'}} - 6 y = 23 sin (3 x)

y^{^{''}} + 4 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

0.125 y^{^{''}} + 106 y + 1000 = 0

What is the solution for y^{''}-5y^'+6y=9x+2e^x ?
The solution for y^{''}-5y^'+6y=9x+2e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+(3x)/2+5/4+e^x