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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

3 y^{^{''}} - 2 y^{^{'}} - 8 y = (2 x + 3) e^{5 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x ^{2}}

y^{^{''}} - 7 y^{^{'}} = 10

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

y^{^{''}} + 9 y = cos (3 t), y (\frac{π}{6}) = 1, y (0) = 1

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