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

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

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y+y^'=sin(t)

y + y^{^{'}} = sin (t)

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

Solution steps

y + y^{'} (t) = sin (t)

Solve linear ODE:

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

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

t^{9} cos (t)

h \to + 0 lim (\frac{∣ h ∣}{h})

\int \frac{ln ( x ) sin ( ln ( x ) )}{x} d x

x \to 1 lim (\frac{x - 1}{x + 1})

\int \frac{e ^{2 x}}{e ^{x} + 3} d x

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