What is the solution for y^{''}-4y^'+5y=10x^2+4x+8 ?

The solution for y^{''}-4y^'+5y=10x^2+4x+8 is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+2x^2+4x+4

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y^{''}-4y^'+5y=10x^2+4x+8

y^{^{''}} - 4 y^{^{'}} + 5 y = 10 x^{2} + 4 x + 8

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = 10 x^{2} + 4 x + 8

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) (e^{- x})

y^{^{''}} + 9 y = 7 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 0.01 y = - 0.01 \cdot 20

y^{^{''}} + \frac{1}{2} y^{^{'}} + y = 4 sin (\frac{x}{2})

y^{^{'''}} - y = x^{3} - 1

What is the solution for y^{''}-4y^'+5y=10x^2+4x+8 ?
The solution for y^{''}-4y^'+5y=10x^2+4x+8 is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+2x^2+4x+4