What is the solution for x^{''}-3x^'+x=-e^t ?

The solution for x^{''}-3x^'+x=-e^t is x=c_{1}e^{((3+sqrt(5))t)/2}+c_{2}e^{((3-sqrt(5))t)/2}+e^t

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x^{''}-3x^'+x=-e^t

x^{^{''}} - 3 x^{^{'}} + x = - e^{t}

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

Solution steps

x^{''} (t) - 3 x^{'} (t) + x = - e^{t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} = 1

y^{^{''}} + 5 y = e^{2 x} - x - x^{2} + x^{3}

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

y^{^{''}} - 2 y^{^{'}} + 2 y = cos^{2} (x)

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 12.5 \cdot sin (2 t)

What is the solution for x^{''}-3x^'+x=-e^t ?
The solution for x^{''}-3x^'+x=-e^t is x=c_{1}e^{((3+sqrt(5))t)/2}+c_{2}e^{((3-sqrt(5))t)/2}+e^t