What is the solution for y^'+sin(t)y=-sin(t),y(0)=9 ?

The solution for y^'+sin(t)y=-sin(t),y(0)=9 is y=10e^{cos(t)-1}-1

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y^'+sin(t)y=-sin(t),y(0)=9

y^{^{'}} + sin (t) y = - sin (t), y (0) = 9

y = 10 e^{c o s (t) - 1} - 1

Solution steps

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

Solve separable ODE:

y = 10 e^{c o s (t) - 1} - 1

y = 10 e^{c o s (t) - 1} - 1

x (x - 1) y + x y^{^{'}} + 2 y = 0

\frac{d}{d x} (\frac{1}{x} + \frac{1}{y}) = 1

\frac{d}{d x} (e^{x} sin (y)) = 0

\frac{d y}{d t} = \frac{t}{t ^{2} y - y}

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

What is the solution for y^'+sin(t)y=-sin(t),y(0)=9 ?
The solution for y^'+sin(t)y=-sin(t),y(0)=9 is y=10e^{cos(t)-1}-1