What is the solution for y^{''}-15y^'+54y=e^{ax} ?

The solution for y^{''}-15y^'+54y=e^{ax} is y=c_{1}e^{9x}+c_{2}e^{6x}+(e^{ax})/(a^2-15a+54)

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y^{''}-15y^'+54y=e^{ax}

y^{^{''}} - 15 y^{^{'}} + 54 y = e^{a x}

y = c_{1} e^{9 x} + c_{2} e^{6 x} + \frac{e ^{a x}}{a ^{2} - 15 a + 54}

Solution steps

y^{''} (x) - 15 y^{'} (x) + 54 y = e^{a x}

Solve linear ODE:

y = c_{1} e^{9 x} + c_{2} e^{6 x} + \frac{e ^{a x}}{a ^{2} - 15 a + 54}

y = c_{1} e^{9 x} + c_{2} e^{6 x} + \frac{e ^{a x}}{a ^{2} - 15 a + 54}

y^{^{'''}} + 9 y^{^{''}} + y^{^{'}} + 9 y = 16 e^{x} - 9

y^{^{''}} + 8 y^{^{'}} + 32 y = 2 e^{2 x}

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

y^{^{''}} - 4 y^{^{'}} + 3 y = x e^{x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3} t ln (t)

What is the solution for y^{''}-15y^'+54y=e^{ax} ?
The solution for y^{''}-15y^'+54y=e^{ax} is y=c_{1}e^{9x}+c_{2}e^{6x}+(e^{ax})/(a^2-15a+54)