What is the solution for y^{''}-8y^'+20y=200x^2-39 ?

The solution for y^{''}-8y^'+20y=200x^2-39 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+10x^2+8x+1/4

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y^{''}-8y^'+20y=200x^2-39

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 39

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 20 y = 200 x^{2} - 39

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = e^{4 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4

y^{^{''}} + 2 y^{^{'}} - 8 y = 3 e^{- 2 x} - e^{- x}

y^{^{''}} - a^{2} y = sec (a x)

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

What is the solution for y^{''}-8y^'+20y=200x^2-39 ?
The solution for y^{''}-8y^'+20y=200x^2-39 is y=e^{4x}(c_{1}cos(2x)+c_{2}sin(2x))+10x^2+8x+1/4