What is the solution for y^{''}-5y^'+4y=(12x-5)e^{4x} ?

The solution for y^{''}-5y^'+4y=(12x-5)e^{4x} is y=c_{1}e^{4x}+c_{2}e^x+2x^2e^{4x}-3e^{4x}x

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y^{''}-5y^'+4y=(12x-5)e^{4x}

y^{^{''}} - 5 y^{^{'}} + 4 y = (12 x - 5) e^{4 x}

y = c_{1} e^{4 x} + c_{2} e^{x} + 2 x^{2} e^{4 x} - 3 e^{4 x} x

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = (12 x - 5) e^{4 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} + 2 x^{2} e^{4 x} - 3 e^{4 x} x

y = c_{1} e^{4 x} + c_{2} e^{x} + 2 x^{2} e^{4 x} - 3 e^{4 x} x

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

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

y^{^{''}} - y = tan (x)

y^{^{''}} + y = 6 x sin (x)

y^{^{''}} + 6 y^{^{'}} + 13 y = 4 x^{2} - 5 x + 2 + 3 e^{x}

What is the solution for y^{''}-5y^'+4y=(12x-5)e^{4x} ?
The solution for y^{''}-5y^'+4y=(12x-5)e^{4x} is y=c_{1}e^{4x}+c_{2}e^x+2x^2e^{4x}-3e^{4x}x