What is the solution for yy^'=8sin(x) ?

The solution for yy^'=8sin(x) is y=sqrt(-16cos(x)+c_{1)},y=-sqrt(-16cos(x)+c_{1)}

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yy^'=8sin(x)

y y^{^{'}} = 8 sin (x)

y = - 16 cos (x) + c_{1}, y = - - 16 cos (x) + c_{1}

Solution steps

y y^{'} (x) = 8 sin (x)

Solve separable ODE:

y = - 16 cos (x) + c_{1}, y = - - 16 cos (x) + c_{1}

y = - 16 cos (x) + c_{1}, y = - - 16 cos (x) + c_{1}

\frac{d}{d x} (\frac{x}{( x - 8 ) ^{2}})

f (x) = e^{x}, at x = ln (12)

x \to 1 lim (x^{3} - x^{2} + 4 x)

sin (2 x + 3)

\frac{d}{d x} ((8 x + 5)^{2})

What is the solution for yy^'=8sin(x) ?
The solution for yy^'=8sin(x) is y=sqrt(-16cos(x)+c_{1)},y=-sqrt(-16cos(x)+c_{1)}