What is the solution for y^{''}-5y^'+6e^x=0 ?

The solution for y^{''}-5y^'+6e^x=0 is y=c_{1}e^{5x}+c_{2}+3/2 e^x

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y^{''}-5y^'+6e^x=0

y^{^{''}} - 5 y^{^{'}} + 6 e^{x} = 0

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 e^{x} = 0

Solve linear ODE:

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

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

3 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = x - 7

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) - 50 cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 4} e^{5 t}

y^{^{''}} - 2 y^{^{'}} + 17 y = 289 x^{2} + 9

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

What is the solution for y^{''}-5y^'+6e^x=0 ?
The solution for y^{''}-5y^'+6e^x=0 is y=c_{1}e^{5x}+c_{2}+3/2 e^x