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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

2 sin (2 x) d x + 3 e^{ey} d y = 2 x d x

\frac{d y}{d x} = 2 y^{2} + 2 y - 12

x \frac{d y}{d x} - 2 y = 0, y = 4 x

(t^{2} + 1) (x + 1) = \frac{d x}{d t} \cdot t

x^{^{'}} + 2 e^{- 4 t} x = 0, x (0) = 5

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