What is the solution for y^{''}-y^'-x-1=0 ?

The solution for y^{''}-y^'-x-1=0 is y=c_{1}e^x+c_{2}-(x^2)/2-2x

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y^{''}-y^'-x-1=0

y^{^{''}} - y^{^{'}} - x - 1 = 0

y = c_{1} e^{x} + c_{2} - \frac{x ^{2}}{2} - 2 x

Solution steps

y^{''} (x) - y^{'} (x) - x - 1 = 0

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{x ^{2}}{2} - 2 x

y = c_{1} e^{x} + c_{2} - \frac{x ^{2}}{2} - 2 x

y^{^{''}} + 6 y^{^{'}} + 25 y = 15 sin (5 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 e^{(- x)} sec^{3} (x)

x, d (x) = x d t (x)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 3 e^{2 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (x) cos (x)

What is the solution for y^{''}-y^'-x-1=0 ?
The solution for y^{''}-y^'-x-1=0 is y=c_{1}e^x+c_{2}-(x^2)/2-2x