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

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

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y^'+ycos(t)=sin(t)cos(t)

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

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

Solution steps

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

Solve linear ODE:

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

\int + \frac{x + 7}{x ^{2} + 4} d x

x^{3} (4 - x)

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 2

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\int 3 sin (x) cos (x) d x

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