What is the solution for y^'+tan(2*x)y=0,y(0)=5 ?

The solution for y^'+tan(2*x)y=0,y(0)=5 is y=cos^{1/2}(2x)*5

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y^'+tan(2*x)y=0,y(0)=5

y^{^{'}} + tan (2 \cdot x) y = 0, y (0) = 5

y = cos^{\frac{1}{2}} (2 x) \cdot 5

Solution steps

y^{'} (x) + tan (2 \cdot x) y = 0

Solve separable ODE:

y = cos^{\frac{1}{2}} (2 x) \cdot 5

y = cos^{\frac{1}{2}} (2 x) \cdot 5

V^{^{'}} = 4 kπ (\frac{3}{4 π})^{\frac{2}{3}} V^{\frac{2}{3}}

y^{^{'}} = \frac{21 x}{y}

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

\frac{d y}{d x} = - 7 x y^{3}

What is the solution for y^'+tan(2*x)y=0,y(0)=5 ?
The solution for y^'+tan(2*x)y=0,y(0)=5 is y=cos^{1/2}(2x)*5