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

The solution for (y^{''}+2y^'+5)=e^{x/3} is y=c_{1}+c_{2}e^{-2x}+9/7 e^{x/3}-(5x)/2

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(y^{''}+2y^'+5)=e^{x/3}

(y^{^{''}} + 2 y^{^{'}} + 5) = e^{\frac{x}{3}}

y = c_{1} + c_{2} e^{- 2 x} + \frac{9}{7} e^{\frac{x}{3}} - \frac{5 x}{2}

Solution steps

(y^{''} (x) + 2 y^{'} (x) + 5) = e^{\frac{x}{3}}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{9}{7} e^{\frac{x}{3}} - \frac{5 x}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{9}{7} e^{\frac{x}{3}} - \frac{5 x}{2}

y^{^{''}} + 5 y = - 180 x^{2} e^{(5 x)}

y^{^{''}} - 10 y^{^{'}} + 21 y = e^{7 t}

y^{^{''}} + 7 y^{^{'}} + 10 = x

y^{^{''}} - 9 y = (x^{2} - 2) \cdot sin (3 x)

y^{^{'''}} - 16 y^{^{'}} = t e^{- 4 t}

What is the solution for (y^{''}+2y^'+5)=e^{x/3} ?
The solution for (y^{''}+2y^'+5)=e^{x/3} is y=c_{1}+c_{2}e^{-2x}+9/7 e^{x/3}-(5x)/2