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

The solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-2 is y=-sqrt(x^2+tan(x)+4)

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(dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-2

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

y = - x^{2} + tan (x) + 4

Solution steps

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

Solve separable ODE:

y = - x^{2} + tan (x) + 4

y = - x^{2} + tan (x) + 4

y^{^{'}} = - 5 y e^{t}, y (0) = 1

2 \frac{d y}{d x} = \frac{x e ^{- y}}{x ^{2} + 3}

t^{2} (1 + y^{2}) + 2 y \frac{d y}{d t} = 0

\frac{d P}{d t} = 0.05 P - 0.00125 P (t)^{2}

y e^{3 x} d x = (9 - e^{3 x}) d y

What is the solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-2 ?
The solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-2 is y=-sqrt(x^2+tan(x)+4)