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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 1

(D^{3} - 3 D^{2} + 4 D - 2) y = cos (x)

\frac{d ^{2} y}{d t ^{2}} - 3 \frac{d y}{d t} + 4 y = cos (2 t)

y^{^{''}} + 5 y^{^{'}} = 100 sin (5 t) + 150 cos (5 t)

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3})

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