What is the solution for y^{''}-13y^'+40y=27+40x ?

The solution for y^{''}-13y^'+40y=27+40x is y=c_{1}e^{8x}+c_{2}e^{5x}+x+1

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y^{''}-13y^'+40y=27+40x

y^{^{''}} - 13 y^{^{'}} + 40 y = 27 + 40 x

y = c_{1} e^{8 x} + c_{2} e^{5 x} + x + 1

Solution steps

y^{''} (x) - 13 y^{'} (x) + 40 y = 27 + 40 x

Solve linear ODE:

y = c_{1} e^{8 x} + c_{2} e^{5 x} + x + 1

y = c_{1} e^{8 x} + c_{2} e^{5 x} + x + 1

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + 1 = 0

(d^{2} + d - 6) y = e^{- 3 x}

3 y^{^{''}} + 5 y^{^{'}} - y = 10

y^{^{''}} + 2 y^{^{'}} - 8 y = 12 e^{2 x}

y^{^{''}} - 5 y^{^{'}} - 6 y = - 3 x^{2}

What is the solution for y^{''}-13y^'+40y=27+40x ?
The solution for y^{''}-13y^'+40y=27+40x is y=c_{1}e^{8x}+c_{2}e^{5x}+x+1