What is the solution for y^{''}-7y^'+10y=e^x ?

The solution for y^{''}-7y^'+10y=e^x is y=c_{1}e^{5x}+c_{2}e^{2x}+1/4 e^x

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

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

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{1}{4} e^{x}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 10 y = e^{x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{1}{4} e^{x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{1}{4} e^{x}

y^{^{''}} + k y = g - 6 t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 8 sin (4 x)

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

y^{^{''}} + y^{^{'}} - 2 y = t e^{3 t}

y^{^{''}} + 24 y^{^{'}} + 144 y = t^{2} e^{- 12 t} + e^{- 12 t}

What is the solution for y^{''}-7y^'+10y=e^x ?
The solution for y^{''}-7y^'+10y=e^x is y=c_{1}e^{5x}+c_{2}e^{2x}+1/4 e^x