What is the solution for y^'=(2x+sec^2(x))/(2y),y(0)=-7 ?

The solution for y^'=(2x+sec^2(x))/(2y),y(0)=-7 is y=-sqrt(x^2+tan(x)+49)

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y^'=(2x+sec^2(x))/(2y),y(0)=-7

y^{^{'}} = \frac{2 x + sec ^{2} ( x )}{2 y}, y (0) = - 7

y = - x^{2} + tan (x) + 49

Solution steps

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

Solve separable ODE:

y = - x^{2} + tan (x) + 49

y = - x^{2} + tan (x) + 49

\frac{d}{d x} (2^{(2 - x^{4})})

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

\int (\frac{( 3 x - 2 ) ^{2} + ( 3 x + 2 ) ^{2}}{x ^{2}}) d x

f (x) = \frac{x}{x + 2}

x y^{^{'}} + y = - 2 x^{6} y^{4}

What is the solution for y^'=(2x+sec^2(x))/(2y),y(0)=-7 ?
The solution for y^'=(2x+sec^2(x))/(2y),y(0)=-7 is y=-sqrt(x^2+tan(x)+49)