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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

x \to 0 lim ((2 x - 1)^{3})

x \to 9 - lim (\frac{- 3}{( x - 9 ) ^{2}})

\int_{- 1}^{1} 1 - y^{2} - (y^{2} - 1) d y

(3, - 2), (4, - 7)

\frac{\partial}{\partial x} (- 3 x e^{- 2 x y})

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