What is the solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-5 ?

The solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-5 is y=-sqrt(x^2+tan(x)+25)

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(dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-5

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{2 y}, y (0) = - 5

y = - x^{2} + tan (x) + 25

Solution steps

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

Solve separable ODE:

y = - x^{2} + tan (x) + 25

y = - x^{2} + tan (x) + 25

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

x y + x y^{^{'}} = x

\frac{d x}{d t} = t x^{2}

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (4) = 4

\frac{d y}{d t} + 2 y = 11, y (0) = 1

What is the solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-5 ?
The solution for (dy)/(dx)=(2x+sec^2(x))/(2y),y(0)=-5 is y=-sqrt(x^2+tan(x)+25)