What is the solution for y^'= y/x+(6x^3)/y cos(x^2) ?

The solution for y^'= y/x+(6x^3)/y cos(x^2) is y=xsqrt(6sin(x^2)+c_{1)},y=-xsqrt(6sin(x^2)+c_{1)}

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y^'= y/x+(6x^3)/y cos(x^2)

y^{^{'}} = \frac{y}{x} + \frac{6 x ^{3}}{y} cos (x^{2})

y = x 6 sin (x^{2}) + c_{1}, y = - x 6 sin (x^{2}) + c_{1}

Solution steps

y^{'} (x) = \frac{y}{x} + \frac{6 x ^{3}}{y} cos (x^{2})

Solve Bernoulli ODE:

y = x 6 sin (x^{2}) + c_{1}, y = - x 6 sin (x^{2}) + c_{1}

y = x 6 sin (x^{2}) + c_{1}, y = - x 6 sin (x^{2}) + c_{1}

y = 2 x + x^{2}, (- 2, 0)

\int x \cdot x d x

y = (x^{2} - 7 x)^{2}

\frac{x}{x ^{2} + 100}

\int_{0}^{2} \frac{1}{x} d x

What is the solution for y^'= y/x+(6x^3)/y cos(x^2) ?
The solution for y^'= y/x+(6x^3)/y cos(x^2) is y=xsqrt(6sin(x^2)+c_{1)},y=-xsqrt(6sin(x^2)+c_{1)}