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

The solution for y^'=ycos(2x),y(0)=1.6 is y=1.6e^{1/2 sin(2x)}

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

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

y = 1.6 e^{\frac{1}{2} s i n (2 x)}

Solution steps

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

Solve separable ODE:

y = 1.6 e^{\frac{1}{2} s i n (2 x)}

y = 1.6 e^{\frac{1}{2} s i n (2 x)}

y = (\frac{x ^{2} + 1}{x ^{2} - 1})^{3}

\int_{1}^{4} 5 - 2 t + 3 t^{2} d t

\frac{\partial}{\partial x} (2 x + 3 y + 4)

\int 2 x (x^{2} + 5)^{4} d x

f (x) = x^{2} (4 - x)^{2}

What is the solution for y^'=ycos(2x),y(0)=1.6 ?
The solution for y^'=ycos(2x),y(0)=1.6 is y=1.6e^{1/2 sin(2x)}