What is the solution for y^{''}-5y^'+4y=8x*e^{2x} ?

The solution for y^{''}-5y^'+4y=8x*e^{2x} is y=c_{1}e^{4x}+c_{2}e^x-4xe^{2x}+2e^{2x}

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y^{''}-5y^'+4y=8x*e^{2x}

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 x \cdot e^{2 x}

y = c_{1} e^{4 x} + c_{2} e^{x} - 4 x e^{2 x} + 2 e^{2 x}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = 8 x \cdot e^{2 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - 4 x e^{2 x} + 2 e^{2 x}

y = c_{1} e^{4 x} + c_{2} e^{x} - 4 x e^{2 x} + 2 e^{2 x}

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

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

y^{^{''}} + y = 8 e^{- 6 x}

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

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-5y^'+4y=8x*e^{2x} ?
The solution for y^{''}-5y^'+4y=8x*e^{2x} is y=c_{1}e^{4x}+c_{2}e^x-4xe^{2x}+2e^{2x}