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

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

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y^{''}-y^'=x+sin(x)

y^{^{''}} - y^{^{'}} = x + sin (x)

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

Solution steps

y^{''} (x) - y^{'} (x) = x + sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 17 y = 5 e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{( 9 + e ^{x} )}

y^{^{''}} + 10 y^{^{'}} + 25 y = 128 e^{3 t}

d, d (x) = lo g_{10} (\frac{x + 4}{x - 1})

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

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