What is the solution for y^'=ysin(2x),y(0)=1.6 ?

The solution for y^'=ysin(2x),y(0)=1.6 is y=1.6e^{-1/2 cos(2x)+1/2}

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y^'=ysin(2x),y(0)=1.6

y^{^{'}} = y sin (2 x), y (0) = 1.6

y = 1.6 e^{- \frac{1}{2} c o s (2 x) + \frac{1}{2}}

Solution steps

y^{'} (x) = y sin (2 x)

Solve separable ODE:

y = 1.6 e^{- \frac{1}{2} c o s (2 x) + \frac{1}{2}}

y = 1.6 e^{- \frac{1}{2} c o s (2 x) + \frac{1}{2}}

\int \frac{16}{x ^{2}} d x

\int_{0}^{4} x^{2} 16 - x^{2} d x

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

n = 0 \sum \infty \frac{1}{2 ^{2 n}}

What is the solution for y^'=ysin(2x),y(0)=1.6 ?
The solution for y^'=ysin(2x),y(0)=1.6 is y=1.6e^{-1/2 cos(2x)+1/2}