What is the solution for y^'=(2x+sec^2(x))/(2y),y(0)=-3 ?

The solution for y^'=(2x+sec^2(x))/(2y),y(0)=-3 is y=-sqrt(x^2+tan(x)+9)

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y^'=(2x+sec^2(x))/(2y),y(0)=-3

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

y = - x^{2} + tan (x) + 9

Solution steps

y^{'} (x) = \frac{2 x + sec ^{2} ( x )}{2 y}

Solve separable ODE:

y = - x^{2} + tan (x) + 9

y = - x^{2} + tan (x) + 9

\frac{\partial}{\partial y} (sin (x y^{2} z^{3}))

\frac{\partial}{\partial x} (6 y^{2} + 284 x - 12)

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

y = e^{7}

y^{^{''}} - 6 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 7

What is the solution for y^'=(2x+sec^2(x))/(2y),y(0)=-3 ?
The solution for y^'=(2x+sec^2(x))/(2y),y(0)=-3 is y=-sqrt(x^2+tan(x)+9)