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

The solution for y^{''}-5y^'+4y=2x^2 is y=c_{1}e^{4x}+c_{2}e^x+(x^2)/2+(5x)/4+21/16

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

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

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{4} + \frac{21}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{4} + \frac{21}{16}

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x ^{2}}{2} + \frac{5 x}{4} + \frac{21}{16}

y^{^{''}} + 16 y = 7 x, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 cos (x)

y^{^{''}} + 4 y = 3 x (2 + cos (2 x))

y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-5y^'+4y=2x^2 ?
The solution for y^{''}-5y^'+4y=2x^2 is y=c_{1}e^{4x}+c_{2}e^x+(x^2)/2+(5x)/4+21/16