What is the solution for y^{''}-7y^'+10y=2xe^x ?

The solution for y^{''}-7y^'+10y=2xe^x is y=c_{1}e^{5x}+c_{2}e^{2x}+(e^xx)/2+5/8 e^x

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y^{''}-7y^'+10y=2xe^x

y^{^{''}} - 7 y^{^{'}} + 10 y = 2 x e^{x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 10 y = 2 x e^{x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

y^{^{''}} - 16 y = 6, y (0) = 1

y^{^{'''}} - 6 y^{^{''}} + 32 y = e^{4 x}

y^{^{''}} + 5 y^{^{'}} + 4 y = 6 e^{- x} + cos (2 x)

y^{^{''}} + 4 y = t^{2} + 8 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} + 49 y^{^{'}} = tan (7 x)

What is the solution for y^{''}-7y^'+10y=2xe^x ?
The solution for y^{''}-7y^'+10y=2xe^x is y=c_{1}e^{5x}+c_{2}e^{2x}+(e^xx)/2+5/8 e^x