What is the solution for y^'+y*cos(t)=-cos(t) ?

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

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y^'+y*cos(t)=-cos(t)

y^{^{'}} + y \cdot cos (t) = - cos (t)

y = e^{- s i n (t) + c_{1}} - 1

Solution steps

y^{'} (t) + y \cdot cos (t) = - cos (t)

Solve separable ODE:

y = e^{- s i n (t) + c_{1}} - 1

y = e^{- s i n (t) + c_{1}} - 1

\frac{d y}{d x} = \frac{ln ( x )}{x y}, y (1) = 7

\frac{d y}{d x} = (x^{2} - 7) y^{2}

\frac{d y}{y} = \frac{d x}{x}

\frac{d y}{d x} = \frac{3 x ^{2} + 1}{2 y - 3}

x^{2} y^{^{'}} = \frac{1}{y} x^{5}

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