What is the solution for y^{''}-8y^'+20y=100x^2-65 ?

The solution for y^{''}-8y^'+20y=100x^2-65 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+5x^2+4x-43/20

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y^{''}-8y^'+20y=100x^2-65

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 65

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 x^{2} + 4 x - \frac{43}{20}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 20 y = 100 x^{2} - 65

Solve linear ODE:

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 x^{2} + 4 x - \frac{43}{20}

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 x^{2} + 4 x - \frac{43}{20}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) \cdot e^{2 x}

y^{^{''}} + 6 y^{^{'}} + 13 y = 10 e^{- 2 t}, y (0) = 3

4 y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 16 y = cos (4 x) + sin (4 x), y^{0} = 3, y^{^{'}} 0 = 5

What is the solution for y^{''}-8y^'+20y=100x^2-65 ?
The solution for y^{''}-8y^'+20y=100x^2-65 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+5x^2+4x-43/20