What is the solution for xy^'+4y=0,y(1)=(sqrt(3))/2 ?

The solution for xy^'+4y=0,y(1)=(sqrt(3))/2 is y=(3^{1/2})/(2x^4)

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xy^'+4y=0,y(1)=(sqrt(3))/2

x y^{^{'}} + 4 y = 0, y (1) = \frac{3}{2}

y = \frac{3 ^{\frac{1}{2}}}{2 x ^{4}}

Solution steps

x y^{'} (x) + 4 y = 0

Solve separable ODE:

y = \frac{3 ^{\frac{1}{2}}}{2 x ^{4}}

y = \frac{3 ^{\frac{1}{2}}}{2 x ^{4}}

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

y^{^{'}} = y \frac{7 ( t ^{2} - 1 )}{t ( t - 3 ) ( t - 4 )}, y (7) = 3

y^{^{'}} = 25 + y^{2}, y (0) = 5 tan (5 x)

\frac{d y}{d x} = \frac{x ^{3} y ^{3}}{x + 3}

y^{^{'}} - 2 x y = 3 x

What is the solution for xy^'+4y=0,y(1)=(sqrt(3))/2 ?
The solution for xy^'+4y=0,y(1)=(sqrt(3))/2 is y=(3^{1/2})/(2x^4)