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

The solution for y^{''}-2y^'+y=x^2-1 is y=c_{1}e^x+c_{2}xe^x+x^2+4x+5

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + x^{2} + 4 x + 5

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x^{2} + 4 x + 5

y = c_{1} e^{x} + c_{2} x e^{x} + x^{2} + 4 x + 5

y^{^{'}} + 4 y = e^{- t} - 4

\frac{d}{d x} (\frac{3}{2 3 x})

y = (5)^{x}, at x = 6

x \to - 2 lim (\frac{- 1}{( x + 2 )})

\frac{d}{d x} ((2 x - 1)^{3} (x + 1)^{5})

What is the solution for y^{''}-2y^'+y=x^2-1 ?
The solution for y^{''}-2y^'+y=x^2-1 is y=c_{1}e^x+c_{2}xe^x+x^2+4x+5