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

The solution for y^'=(2x+sec^2(x))/(2y) is y=sqrt(x^2+tan(x)+c_{1)},y=-sqrt(x^2+tan(x)+c_{1)}

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y^'=(2x+sec^2(x))/(2y)

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

y = x^{2} + tan (x) + c_{1}, y = - x^{2} + tan (x) + c_{1}

Solution steps

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

Solve separable ODE:

y = x^{2} + tan (x) + c_{1}, y = - x^{2} + tan (x) + c_{1}

y = x^{2} + tan (x) + c_{1}, y = - x^{2} + tan (x) + c_{1}

x^{2} sin (3 x)

2 x sin (y) d x + x^{2} cos (y) d y = 0

\int - 24 x^{5} d x

\int (x^{2} + sec^{2} (x)) d x

n = 0 \sum \infty 0.7 4^{n}

What is the solution for y^'=(2x+sec^2(x))/(2y) ?
The solution for y^'=(2x+sec^2(x))/(2y) is y=sqrt(x^2+tan(x)+c_{1)},y=-sqrt(x^2+tan(x)+c_{1)}