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

The solution for y^{''}-4y^'+5y=5x^2+2x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x^2+2x+6/5

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

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

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

Solution steps

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

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

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

y^{^{''}} - y^{^{'}} + 4 y = 5 t^{2} + 9 t + 4

y^{^{'''}} + 3 y^{^{'}} + 2 = 0

y^{^{''}} + 4 y^{^{'}} + 12 y = 80 sin (2 x)

y^{^{''}} - y^{^{'}} + y = 3 sin (t) - 27 e^{2 t}

What is the solution for y^{''}-4y^'+5y=5x^2+2x ?
The solution for y^{''}-4y^'+5y=5x^2+2x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x^2+2x+6/5