What is the solution for y^{''}-2y^'+y=(3+5x)e^{2x} ?

The solution for y^{''}-2y^'+y=(3+5x)e^{2x} is y=c_{1}e^x+c_{2}xe^x+5xe^{2x}-7e^{2x}

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y^{''}-2y^'+y=(3+5x)e^{2x}

y^{^{''}} - 2 y^{^{'}} + y = (3 + 5 x) e^{2 x}

y = c_{1} e^{x} + c_{2} x e^{x} + 5 x e^{2 x} - 7 e^{2 x}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = (3 + 5 x) e^{2 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 5 x e^{2 x} - 7 e^{2 x}

y = c_{1} e^{x} + c_{2} x e^{x} + 5 x e^{2 x} - 7 e^{2 x}

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 2, y^{^{''}} (0) = - 1

y^{^{''}} - 2 y^{^{'}} + 2 y = t cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{- 1}{2}} \cdot e^{x}

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 7 x

What is the solution for y^{''}-2y^'+y=(3+5x)e^{2x} ?
The solution for y^{''}-2y^'+y=(3+5x)e^{2x} is y=c_{1}e^x+c_{2}xe^x+5xe^{2x}-7e^{2x}