What is the solution for y^{''}+3y^'+2y=10e^x ?

The solution for y^{''}+3y^'+2y=10e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+5/3 e^x

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y^{''}+3y^'+2y=10e^x

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 e^{x}

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 10 e^{x}

Solve linear ODE:

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

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

(d^{3} + 2 d^{2} - 5 d - 6) y = 6^{- 3} x

y^{^{''}} + 8 y^{^{'}} + 15 y = 10, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 36 y = 6 e^{2} x

- x^{^{''}} + x = 4

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

What is the solution for y^{''}+3y^'+2y=10e^x ?
The solution for y^{''}+3y^'+2y=10e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+5/3 e^x