What is the solution for y^{''}-8y^'+20y=200x^2-91 ?

The solution for y^{''}-8y^'+20y=200x^2-91 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+10x^2+8x-47/20

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y^{''}-8y^'+20y=200x^2-91

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 91

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 10 x^{2} + 8 x - \frac{47}{20}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 20 y = 200 x^{2} - 91

Solve linear ODE:

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 10 x^{2} + 8 x - \frac{47}{20}

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 10 x^{2} + 8 x - \frac{47}{20}

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

x^{^{''}} + 16 x = cos (4 t), x (0) = 0, x^{^{'}} (0) = 1

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{4} e^{t}

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

(D^{2} - 1) y = x e^{3 x}

What is the solution for y^{''}-8y^'+20y=200x^2-91 ?
The solution for y^{''}-8y^'+20y=200x^2-91 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+10x^2+8x-47/20