What is the solution for y^{''}-5y^'+6y=2e^x ?

The solution for y^{''}-5y^'+6y=2e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+e^x

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y^{''}-5y^'+6y=2e^x

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + e^{x}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 2 e^{x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + e^{x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + e^{x}

(d^{2} - 1) \cdot y = x \cdot sin (x) + x^{2} \cdot e^{x}

d, d (x) = \frac{1}{36 ( x - 60 ) ^{2} - 100}

(d^{2} + 2 d + 1) \cdot y = x \cdot e^{- x} \cdot cos (x)

(d^{2} + 5 d + 4) y = sin^{3} (x)

(d^{3} - 7 d + 6) \cdot y = sin^{2} (x)

What is the solution for y^{''}-5y^'+6y=2e^x ?
The solution for y^{''}-5y^'+6y=2e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+e^x