What is the solution for y^{''}-2y^'+10y=15x^2-x+1 ?

The solution for y^{''}-2y^'+10y=15x^2-x+1 is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+(3x^2)/2+x/2-1/10

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y^{''}-2y^'+10y=15x^2-x+1

y^{^{''}} - 2 y^{^{'}} + 10 y = 15 x^{2} - x + 1

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3 x ^{2}}{2} + \frac{x}{2} - \frac{1}{10}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 10 y = 15 x^{2} - x + 1

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3 x ^{2}}{2} + \frac{x}{2} - \frac{1}{10}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3 x ^{2}}{2} + \frac{x}{2} - \frac{1}{10}

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

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

y^{^{''}} - 17 y^{^{'}} + 72 y = 55 + 72 x

y^{^{''}} - y^{^{'}} - 6 y = 5 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 7 y = 6 cos (2 x)

What is the solution for y^{''}-2y^'+10y=15x^2-x+1 ?
The solution for y^{''}-2y^'+10y=15x^2-x+1 is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+(3x^2)/2+x/2-1/10