What is the solution for y^{''}-3y^'+y=x ?

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

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

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

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + x + 3

Solution steps

y^{''} (x) - 3 y^{'} (x) + y = x

Solve linear ODE:

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + x + 3

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + x + 3

y^{^{''}} + y^{^{'}} = x (e^{3 x} + 1)

f^{^{''}} (t) + 9 \cdot f (t) = (- 2 \cdot e^{(3 t)})

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 3 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = - 432 t^{3}

y^{^{''}} + v y^{^{'}} = c e^{(- r^{2})}

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