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

The solution for xy^'=y+2x^3sin^2(y/x) is y=arccot(-x^2-c_{1})x+pinx

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xy^'=y+2x^3sin^2(y/x)

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

y = arccot (- x^{2} - c_{1}) x + πn x

Solution steps

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

Solve using substitution:

y = arccot (- x^{2} - c_{1}) x + πn x

y = arccot (- x^{2} - c_{1}) x + πn x

\frac{d}{d x} (x^{2} + 100)

\int cos (2 x^{2}) d x

\frac{d}{d x} (\frac{1}{1})

\frac{( y - 2 ) d y}{y + 3} = \frac{( x - 1 ) d x}{x + 4}

ln (6 x^{3} + 3 x^{2} + 3 x + 3)

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