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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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