What is the solution for x^'=-x+sin(t) ?

The solution for x^'=-x+sin(t) is x=-(cos(t))/2+(sin(t))/2+(c_{1})/(e^t)

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x^'=-x+sin(t)

x^{^{'}} = - x + sin (t)

x = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

Solution steps

x^{'} (t) = - x + sin (t)

Solve linear ODE:

x = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

x = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

What is the solution for x^'=-x+sin(t) ?
The solution for x^'=-x+sin(t) is x=-(cos(t))/2+(sin(t))/2+(c_{1})/(e^t)