What is the solution for (dy)/(dx)-2cos(x)y=0 ?

The solution for (dy)/(dx)-2cos(x)y=0 is y=e^{2sin(x)+c_{1}}

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(dy)/(dx)-2cos(x)y=0

\frac{d y}{d x} - 2 cos (x) y = 0

y = e^{2 s i n (x) + c_{1}}

Solution steps

\frac{d y}{d x} - 2 cos (x) y = 0

Solve separable ODE:

y = e^{2 s i n (x) + c_{1}}

y = e^{2 s i n (x) + c_{1}}

y^{\frac{1}{2}} y^{^{'}} = 1 - y^{\frac{3}{2}}

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

\frac{d y}{d x} = \frac{( 5 x ^{6} )}{y ^{2}}, y (0) = 1

e^{x^{2}} \frac{d y}{d x} + e^{x^{2}} 2 x y = e^{x^{2}} 2 x

y^{^{'}} = - 2 \cdot t \cdot y^{2}, y (0) = 1

What is the solution for (dy)/(dx)-2cos(x)y=0 ?
The solution for (dy)/(dx)-2cos(x)y=0 is y=e^{2sin(x)+c_{1}}